<rt id="ucqwu"></rt>
嚴格按照BP網絡計算公式來設計的一個matlab程序,對BP網絡進行了優化設計 優化1：設計了yyy，即在o(k)計算公式時，當網絡進入平坦區時(<0.0001)學習率加大，出來后學習率又還原 優化2：v(i,j)=v(i,j)+deltv(i,j)+a*dv(i,j)
上傳時間： 2014-11-30
上傳用戶：妄想演繹師
實驗源代碼 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("請輸入矩陣第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可傳遞閉包關系矩陣是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元關系的可傳遞閉包\n"); void warshall(int,int); int k , n; printf("請輸入矩陣的行數 i: "); scanf("%d",&k); 四川大學實驗報告 printf("請輸入矩陣的列數 j: "); scanf("%d",&n); warshall(k,n); }
上傳時間： 2016-06-27
上傳用戶：梁雪文以
#include "STC90.h" #include < intrins.h > #define uchar unsigned char #define uint unsigned int #define led_port P1 sbit IR_RE = P3^2; sbit led_r = P1^3; sbit led_g = P1^4; sbit led_b = P1^5; sbit led_wd = P1^7; sbit K1 =P3^0 ; //增加鍵 sbit K2 =P3^1 ; //減少鍵 sbit BEEP =P3^7 ; //蜂鳴器 uchar temp,temp1; bit k=0; //紅外解碼判斷標志位，為0則為有效信號，為1則為無效 bit Flag2; uchar date[4]={0,0,0,0}; //date數組為存放地址原碼，反碼，數據原碼，反碼 uint lade_1,lade_2,lade_3,lade_4; uint num; uchar date_ram,ee_temp,ee_temp1; uchar WDT_NUM=0; uchar const dofly[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f};// 顯示段碼值01234567 uchar code seg[]={7,6,5,4,3,2,1,0};//分別對應相應的數碼管點亮,即位碼 unsigned long disp_date; void fade(); void fade1(); /*************************** 看門狗子程序*************************/ void watchdog_timer() { if(WDT_NUM==5) { WDT_NUM=0; led_wd=!led_wd; } WDT_NUM++; WDT_CONTR=0x3f; } /******************************************************************/ void delay(unsigned int cnt) { while(--cnt); } /*--------------------------延時1ms程子程序-----------------------*/ void delay_1ms(uint z) { uint x,y; for(x=z;x>0;x--) for(y=126;y>0;y--); } /*--------------------------延時1ms程子程序-----------------------*/ delay1000() { uchar i,j; i=5; do{j=95; do{j--;} while(j); i--; } while(i); } /*---------------------------延時882us子程序-----------------------*/ delay882() { uchar i,j; i=6; do{j=71; do{j--;} while(j); i--; }while(i); } /*--------------------------延時2400us程子程序-----------------------*/ delay2400() { uchar i,j; i=5; do{j=237; do{j--;} while(j); i--; }while(i); } /**********************************************************************/ /* void display() { uchar i; for(i=0;i<8;i++) { P0=dofly[disp_date%10];//取顯示數據，段碼 P2=seg[i]; //取位碼 delay_1ms(1); disp_date/=10; } } */ /*********************************************************************/ uchar EEPROM_read(uint addr)//EEPROM字節讀 { ISP_CONTR=0x83; //系統時鐘<12M時，對ISP_CONTR寄存器設置的值，本電路為11.0592M ISP_CMD=1; //字節讀 ISP_ADDRH=(addr&0xff00)>>8; ISP_ADDRL=addr&0x00ff; ISP_TRIG=0x46; ISP_TRIG=0xb9; _nop_(); _nop_(); return ISP_DATA; } //-------------------------------------------------------------------- void EEPROM_write(uint addr,uchar dat)//EEPROM字節寫 { ISP_CONTR=0x83; //系統時鐘<12M時，對ISP_CONTR寄存器設置的值，本電路為11.0592M ISP_CMD=2; //字節編程 ISP_ADDRH=(addr&0xff00)>>8; ISP_ADDRL=addr&0x00ff; ISP_DATA=dat; ISP_TRIG=0x46; ISP_TRIG=0xb9; _nop_(); _nop_(); } //-------------------------------------------------------------------- void EEPROM_ERASE(uint addr)//EEPROM扇區擦除 { ISP_CONTR=0x83; //系統時鐘<12M時，對ISP_CONTR寄存器設置的值，本電路為11.0592M ISP_CMD=3; //扇區擦除 ISP_ADDRH=(addr&0xff00)>>8; ISP_ADDRL=addr&0x00ff; ISP_TRIG=0x46; ISP_TRIG=0xb9; _nop_(); _nop_(); } //************************************************************** /*----------------------------------------------------------*/ /*-----------------------紅外解碼程序(核心)-----------------*/ /*----------------------------------------------------------*/ void IR_decode() { uchar i,j; while(IR_RE==0); delay2400(); if(IR_RE==1) //延時2.4ms后如果是高電平則是新碼 { delay1000(); delay1000(); for(i=0;i<4;i++) { for(j=0;j<8;j++) { while(IR_RE==0); //等待地址碼第1位高電平到來 delay882(); //延時882us判斷此時引腳電平 ///CY=IR_RE; if(IR_RE==0) { date[i]>>=1; date[i]=date[i]|0x00; } else if(IR_RE==1) { delay1000(); date[i]>>=1; date[i]=date[i]|0x80; } } //1位數據接收結束 } //32位二進制碼接收結束 } } /* void LED_PWM() { lade_2=num; //384 lade_4=num; //384 while(lade_2!=0&Flag2==1) { for(lade_3=512;lade_3>lade_4;lade_3--) //512 { led_port=0x00; delay(1); } lade_3=512; //512 lade_4--; for(lade_1=0;lade_1<lade_2;lade_1++) { led_port=0x38; //c7 delay(1); } lade_1=0; lade_2--; if(temp!=0x0c&Flag2==1) { lade_2=0; } lade_2=num; //384 lade_4=num; //384 } } */ void calc() { EEPROM_read(0x2000); ee_temp1=ISP_DATA; ee_temp=ee_temp1&0x0f; //************************************* 1 /* if(date[3]==0xff&Flag2==1) { if(num>=20) { num=num-80; } //else num=1; LED_PWM(); } if(date[3]==0xfe&Flag2==1) { if(num<=500) { num=num+80; } // else num=511; LED_PWM(); } if(ee_temp1==0xfd) { led_port=0x00; watchdog_timer(); } if(ee_temp1==0xfc) { led_port=0x00; led_r=1; led_g=1; led_b=1; watchdog_timer(); } */ //********************************************** 2 if(ee_temp1==0xfb) { led_port=0x00; led_r=1; watchdog_timer(); } if(ee_temp1==0xfa) { led_port=0x00; led_g=1; watchdog_timer(); } if(ee_temp1==0xf9) { led_port=0x00; led_b=1; watchdog_timer(); } if(ee_temp1==0xf8) { led_port=0x00; led_r=1; led_g=1; led_b=1; watchdog_timer(); } //************************************** 3 if(ee_temp1==0xf7) { uint fade_1,fade_2,fade_3,fade_4; fade_2=448; //384 fade_4=448; //384 while(fade_2!=0&ee_temp==0x07) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x10; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x08; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x07) { fade_2=0; } watchdog_timer(); fade_2=448; //384 fade_4=448; //384 } } if(ee_temp1==0xf6) { uint fade_1,fade_2,fade_3,fade_4; fade_2=448; //384 fade_4=448; //384 while(fade_2!=0&ee_temp==0x06) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x20; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x10; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x06) { fade_2=0; } watchdog_timer(); fade_2=448; //384 fade_4=448; //384 } } if(ee_temp1==0xf5) { uint fade_1,fade_2,fade_3,fade_4; fade_2=448; //384 fade_4=448; //384 while(fade_2!=0&ee_temp==0x05) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x08; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x20; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x05) { fade_2=0; } watchdog_timer(); fade_2=448; //384 fade_4=448; //384 } } if(ee_temp1==0xf4) { while(ee_temp==4) { led_port=0x00; led_r=1; delay_1ms(200); led_port=0x00; led_r=1; led_g=1; delay_1ms(200); led_port=0x00; led_g=1; delay_1ms(200); watchdog_timer(); led_port=0x00; led_g=1; led_b=1; delay_1ms(200); led_port=0x00; led_b=1; delay_1ms(200); led_port=0x00; led_b=1; led_r=1; delay_1ms(200); watchdog_timer(); } } //************************************** 4 if(ee_temp1==0xf3) { uint fade_1,fade_2,fade_3,fade_4; fade_2=416; //384 fade_4=416; //384 while(fade_2!=0&ee_temp==0x03) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x10; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x08; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x03) { fade_2=0; } watchdog_timer(); fade_2=416; //384 fade_4=416; //384 } } if(ee_temp1==0xf2) { uint fade_1,fade_2,fade_3,fade_4; fade_2=384; //384 fade_4=384; //384 while(fade_2!=0&ee_temp==0x02) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x20; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x10; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x02) { fade_2=0; } watchdog_timer(); fade_2=384; //384 fade_4=384; //384 } } if(ee_temp1==0xf1) { uint fade_1,fade_2,fade_3,fade_4; fade_2=348; //384 fade_4=348; //384 while(fade_2!=0&ee_temp==0x01) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x08; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x20; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x01) { fade_2=0; } watchdog_timer(); fade_2=348; //384 fade_4=348; //384 } } if(ee_temp1==0xf0) { while(ee_temp==0) { led_port=0x00; led_r=1; delay_1ms(500); watchdog_timer(); led_port=0x00; led_g=1; delay_1ms(500); led_port=0x00; led_b=1; delay_1ms(500); watchdog_timer(); } } //******************************************** 5 if(ee_temp1==0xef) { uint fade_1,fade_2,fade_3,fade_4; fade_2=384; //384 fade_4=384; //384 while(fade_2!=0&ee_temp==0x0f) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x10; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x08; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x0f) { fade_2=0; } watchdog_timer(); fade_2=384; //384 fade_4=384; //384 } } if(ee_temp1==0xee) { uint fade_1,fade_2,fade_3,fade_4; fade_2=320; //384 fade_4=320; //384 while(fade_2!=0&ee_temp==0x0e) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x20; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x10; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x0e) { fade_2=0; } watchdog_timer(); fade_2=320; //384 fade_4=320; //384 } } if(ee_temp1==0xed) { uint fade_1,fade_2,fade_3,fade_4; fade_2=320; //384 fade_4=320; //384 while(fade_2!=0&ee_temp==0x0d) { for(fade_3=512;fade_3>fade_4;fade_3--) //512 { led_port=0x08; delay(1); } fade_3=512; //512 fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x20; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x0d) { fade_2=0; } watchdog_timer(); fade_2=320; //384 fade_4=320; //384 } } if(ee_temp1==0xec) fade(); //******************************************* 6 if(ee_temp1==0xeb) { led_port=0x00; led_r=1; led_g=1; watchdog_timer(); } if(ee_temp1==0xea) { led_port=0x00; //led_r=0; led_g=1; led_b=1; watchdog_timer(); } if(ee_temp1==0xe9) { led_port=0x00; led_r=1; //led_g=0; led_b=1; watchdog_timer(); } if(ee_temp1==0xe8) fade1(); } void fade() { // uchar i; uint fade_1,fade_2,fade_3,fade_4; fade_2=512; fade_4=511; while(fade_2!=0&ee_temp==0x0c) { for(fade_3=512;fade_3>fade_4;fade_3--) { led_port=0x10; delay(1); } fade_3=512; fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x08; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x0c) { fade_2=0; } } watchdog_timer(); fade_2=512; fade_4=511; while(fade_2!=0&ee_temp==0x0c) { if(ee_temp!=0x0c) { fade_2=0; } for(fade_3=512;fade_3>fade_4;fade_3--) { led_port=0x20; delay(1); // watchdog_timer(); } fade_3=512; fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x10; delay(1); // watchdog_timer(); } fade_1=0; fade_2--; } watchdog_timer(); fade_2=512; fade_4=511; while(fade_2!=0&ee_temp==0x0c) { if(ee_temp!=0x0c) { fade_2=0; } for(fade_3=512;fade_3>fade_4;fade_3--) { led_port=0x08; delay(1); watchdog_timer(); } fade_3=512; fade_4--; watchdog_timer(); for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x20; delay(1); watchdog_timer(); } fade_1=0; fade_2--; } watchdog_timer(); } void fade1() { // uchar i; uint fade_1,fade_2,fade_3,fade_4; fade_2=128; fade_4=127; while(fade_2!=0&ee_temp==0x08) { for(fade_3=128;fade_3>fade_4;fade_3--) { led_port=0x10; delay(1); } fade_3=128; fade_4--; for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x08; delay(1); } fade_1=0; fade_2--; if(ee_temp!=0x08) { fade_2=0; } } watchdog_timer(); fade_2=128; fade_4=127; while(fade_2!=0&ee_temp==0x08) { if(ee_temp!=0x08) { fade_2=0; } for(fade_3=128;fade_3>fade_4;fade_3--) { led_port=0x20; delay(1); } fade_3=128; fade_4--; for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x10; delay(1); } fade_1=0; fade_2--; } watchdog_timer(); fade_2=128; fade_4=127; while(fade_2!=0&ee_temp==0x08) { if(ee_temp!=0x08) { fade_2=0; } for(fade_3=128;fade_3>fade_4;fade_3--) { led_port=0x08; delay(1); } fade_3=128; fade_4--; for(fade_1=0;fade_1<fade_2;fade_1++) { led_port=0x20; delay(1); } fade_1=0; fade_2--; } watchdog_timer(); } void init() { led_port=0x00; /* led_r=1; delay_1ms(500); led_port=0x00; led_g=1; delay_1ms(500); led_port=0x00; led_b=1; delay_1ms(500); led_port=0x00; */ delay_1ms(2); WDT_CONTR=0x3f; delay_1ms(500); } //******************************** void main() { init(); Flag2=0; SP=0x60; //堆棧指針 EX0=1; //允許外部中斷0,用于檢測紅外遙控器按鍵 EA=1; num=255; while(1) { calc(); } } //******************************************************************** /*------------------------外部中斷0程序-------------------------*/ /*------------------主要用于處理紅外遙控鍵值--------------------*/ void int0() interrupt 0 { uchar i; Flag2=0; /////// k=0; EX0=0; //檢測到有效信號關中斷，防止干擾 for(i=0;i<4;i++) { delay1000(); if(IR_RE==1){k=1;} //剛開始為9ms的引導碼. } led_port=0x00; if(k==0) { IR_decode(); //如果接收到的是有效信號，則調用解碼程序 if(date[3]>=0xe8) { if(date[3]<=0xfb) { temp1=date[3]; EEPROM_ERASE(0x2000); //STC_EEROM_0X2000 temp1 EEPROM_write(0x2000,temp1); EEPROM_read(0x2000); ee_temp1=ISP_DATA; ee_temp=ee_temp1&0x0f; /* temp=date[3]&0x0f; EEPROM_ERASE(0x2004); //STC_EEROM_0X2004 temp EEPROM_write(0x2004,temp); */ } else { EEPROM_read(0x2000); ee_temp1=ISP_DATA; ee_temp=ee_temp1&0x0f; } } delay2400(); delay2400(); delay2400(); delay_1ms(500); } EX0=1; //開外部中斷，允許新的遙控按鍵 }
上傳時間： 2016-07-02
上傳用戶：184890962
Computation of loudness (Zwicker model) according to ISO 532B / DIN 45631 norms. This model is valid for steady sounds. Code based on BASIC program published in the following article: Program for calculating loudness according to DIN 45 631 (ISO 532B)", E.Zwicker and H.Fastl, J.A.S.J (E) 12, 1 (1991).
上傳時間： 2016-11-14
上傳用戶：zztony16
void DFS(MGraph G, int i) { int j; visited[i] = TRUE; printf("%c ", G.vexs[i]); for (j=0; j<G.numVertexes; ++j) { if (G.arc[i][j]!=INFINITY && !visited[j]) DFS(G, j); } }
上傳時間： 2016-12-28
上傳用戶：chenyameng
#include <malloc.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #define NULL 0 #define MaxSize 30 typedef struct athletestruct /*運動員*/ { char name[20]; int score; /*分數*/ int range; /**/ int item; /*項目*/ }ATH; typedef struct schoolstruct /*學校*/ { int count; /*編號*/ int serial; /**/ int menscore; /*男選手分數*/ int womenscore; /*女選手分數*/ int totalscore; /*總分*/ ATH athlete[MaxSize]; /**/ struct schoolstruct *next; }SCH; int nsc,msp,wsp; int ntsp; int i,j; int overgame; int serial,range; int n; SCH *head,*pfirst,*psecond; int *phead=NULL,*pafirst=NULL,*pasecond=NULL; void create(); void input () { char answer; head = (SCH *)malloc(sizeof(SCH)); /**/ head->next = NULL; pfirst = head; answer = 'y'; while ( answer == 'y' ) { Is_Game_DoMain: printf("\nGET Top 5 when odd\nGET Top 3 when even"); printf("\n輸入運動項目序號 (x<=%d):",ntsp); scanf("%d",pafirst); overgame = *pafirst; if ( pafirst != phead ) { for ( pasecond = phead ; pasecond < pafirst ; pasecond ++ ) { if ( overgame == *pasecond ) { printf("\n這個項目已經存在請選擇其他的數字\n"); goto Is_Game_DoMain; } } } pafirst = pafirst + 1; if ( overgame > ntsp ) { printf("\n項目不存在"); printf("\n請重新輸入"); goto Is_Game_DoMain; } switch ( overgame%2 ) { case 0: n = 3;break; case 1: n = 5;break; } for ( i = 1 ; i <= n ; i++ ) { Is_Serial_DoMain: printf("\n輸入序號 of the NO.%d (0<x<=%d): ",i,nsc); scanf("%d",&serial); if ( serial > nsc ) { printf("\n超過學校數目,請重新輸入"); goto Is_Serial_DoMain; } if ( head->next == NULL ) { create(); } psecond = head->next ; while ( psecond != NULL ) { if ( psecond->serial == serial ) { pfirst = psecond; pfirst->count = pfirst->count + 1; goto Store_Data; } else { psecond = psecond->next; } } create(); Store_Data: pfirst->athlete[pfirst->count].item = overgame; pfirst->athlete[pfirst->count].range = i; pfirst->serial = serial; printf("Input name:) : "); scanf("%s",pfirst->athlete[pfirst->count].name); } printf("\n繼續輸入運動項目(y&n)？"); answer = getchar(); printf("\n"); } } void calculate() /**/ { pfirst = head->next; while ( pfirst->next != NULL ) { for (i=1;i<=pfirst->count;i++) { if ( pfirst->athlete[i].item % 2 == 0 ) { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 5;break; case 2:pfirst->athlete[i].score = 3;break; case 3:pfirst->athlete[i].score = 2;break; } } else { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 7;break; case 2:pfirst->athlete[i].score = 5;break; case 3:pfirst->athlete[i].score = 3;break; case 4:pfirst->athlete[i].score = 2;break; case 5:pfirst->athlete[i].score = 1;break; } } if ( pfirst->athlete[i].item <=msp ) { pfirst->menscore = pfirst->menscore + pfirst->athlete[i].score; } else { pfirst->womenscore = pfirst->womenscore + pfirst->athlete[i].score; } } pfirst->totalscore = pfirst->menscore + pfirst->womenscore; pfirst = pfirst->next; } } void output() { pfirst = head->next; psecond = head->next; while ( pfirst->next != NULL ) { // clrscr(); printf("\n第%d號學校的結果成績:",pfirst->serial); printf("\n\n項目的數目\t學校的名字\t分數"); for (i=1;i<=ntsp;i++) { for (j=1;j<=pfirst->count;j++) { if ( pfirst->athlete[j].item == i ) { printf("\n %d\t\t\t\t\t\t%s\n %d",i,pfirst->athlete[j].name,pfirst->athlete[j].score);break; } } } printf("\n\n\n\t\t\t\t\t\t按任意建 進入下一頁"); getchar(); pfirst = pfirst->next; } // clrscr(); printf("\n運動會結果:\n\n學校編號\t男運動員成績\t女運動員成績\t總分"); pfirst = head->next; while ( pfirst->next != NULL ) { printf("\n %d\t\t %d\t\t %d\t\t %d",pfirst->serial,pfirst->menscore,pfirst->womenscore,pfirst->totalscore); pfirst = pfirst->next; } printf("\n\n\n\t\t\t\t\t\t\t按任意建結束"); getchar(); } void create() { pfirst = (struct schoolstruct *)malloc(sizeof(struct schoolstruct)); pfirst->next = head->next ; head->next = pfirst ; pfirst->count = 1; pfirst->menscore = 0; pfirst->womenscore = 0; pfirst->totalscore = 0; } void Save() {FILE *fp; if((fp = fopen("school.dat","wb"))==NULL) {printf("can't open school.dat\n"); fclose(fp); return; } fwrite(pfirst,sizeof(SCH),10,fp); fclose(fp); printf("文件已經成功保存\n"); } void main() { system("cls"); printf("\n\t\t\t 運動會分數統計\n"); printf("輸入學校數目 (x>= 5):"); scanf("%d",&nsc); printf("輸入男選手的項目(x<=20):"); scanf("%d",&msp); printf("輸入女選手項目(<=20):"); scanf("%d",&wsp); ntsp = msp + wsp; phead = (int *)calloc(ntsp,sizeof(int)); pafirst = phead; pasecond = phead; input(); calculate(); output(); Save(); }
標簽： 源代碼
上傳時間： 2016-12-28
上傳用戶：150501
1．Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose sum is exactly x. (Implement exercise 2.3-7.) #include<stdio.h> #include<stdlib.h> void merge(int arr[],int low,int mid,int high){ int i,k; int *tmp=(int*)malloc((high-low+1)*sizeof(int)); int left_low=low; int left_high=mid; int right_low=mid+1; int right_high=high; for(k=0;left_low<=left_high&&right_low<=right_high;k++) { if(arr[left_low]<=arr[right_low]){ tmp[k]=arr[left_low++]; } else{ tmp[k]=arr[right_low++]; } } if(left_low<=left_high){ for(i=left_low;i<=left_high;i++){ tmp[k++]=arr[i]; } } if(right_low<=right_high){ for(i=right_low;i<=right_high;i++) tmp[k++]=arr[i]; } for(i=0;i<high-low+1;i++) arr[low+i]=tmp[i]; } void merge_sort(int a[],int p,int r){ int q; if(p<r){ q=(p+r)/2; merge_sort(a,p,q); merge_sort(a,q+1,r); merge(a,p,q,r); } } int main(){ int a[8]={3,5,8,6,4,1,1}; int i,j; int x=10; merge_sort(a,0,6); printf("after Merging-Sort:\n"); for(i=0;i<7;i++){ printf("%d",a[i]); } printf("\n"); i=0;j=6; do{ if(a[i]+a[j]==x){ printf("exist"); break; } if(a[i]+a[j]>x) j--; if(a[i]+a[j]<x) i++; }while(i<=j); if(i>j) printf("not exist"); system("pause"); return 0; }
上傳時間： 2017-04-01
上傳用戶：糖兒水嘻嘻
題目：古典問題：有一對兔子，從出生后第3個月起每個月都生一對兔子，小兔子長到第三個月后每個月又生一對兔子，假如兔子都不死，問每個月的兔子總數為多少？ //這是一個菲波拉契數列問題 public class lianxi01 { public static void main(String[] args) { System.out.println("第1個月的兔子對數: 1"); System.out.println("第2個月的兔子對數: 1"); int f1 = 1, f2 = 1, f, M=24; for(int i=3; i<=M; i++) { f = f2; f2 = f1 + f2; f1 = f; System.out.println("第" + i +"個月的兔子對數: "+f2); } } } 【程序2】 題目：判斷101-200之間有多少個素數，并輸出所有素數。 程序分析：判斷素數的方法：用一個數分別去除2到sqrt(這個數)，如果能被整除， 則表明此數不是素數，反之是素數。 public class lianxi02 { public static void main(String[] args) { int count = 0; for(int i=101; i<200; i+=2) { boolean b = false; for(int j=2; j<=Math.sqrt(i); j++) { if(i % j == 0) { b = false; break; } else { b = true; } } if(b == true) {count ++;System.out.println(i );} } System.out.println( "素數個數是: " + count); } } 【程序3】 題目：打印出所有的 "水仙花數 "，所謂 "水仙花數 "是指一個三位數，其各位數字立方和等于該數本身。例如：153是一個 "水仙花數 "，因為153=1的三次方＋5的三次方＋3的三次方。 public class lianxi03 { public static void main(String[] args) { int b1, b2, b3;
上傳時間： 2017-12-24
上傳用戶：Ariza
#include "iostream" using namespace std; class Matrix { private: double** A; //矩陣A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //為向量b分配空間并初始化為0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //為向量A分配空間并初始化為0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析構中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"請輸入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"請輸入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"個:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分別求得U，L的第一行與第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分別求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"計算U得:"<<endl; U.Disp(); cout<<"計算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
標簽： 道理特分解法
上傳時間： 2018-05-20
上傳用戶：Aa123456789
function [alpha,N,U]=youxianchafen2(r1,r2,up,under,num,deta) %[alpha,N,U]=youxianchafen2(a,r1,r2,up,under,num,deta) %該函數用有限差分法求解有兩種介質的正方形區域的二維拉普拉斯方程的數值解 %函數返回迭代因子、迭代次數以及迭代完成后所求區域內網格節點處的值 %a為正方形求解區域的邊長 %r1,r2分別表示兩種介質的電導率 %up,under分別為上下邊界值 %num表示將區域每邊的網格剖分個數 %deta為迭代過程中所允許的相對誤差限 n=num+1; %每邊節點數 U(n,n)=0; %節點處數值矩陣 N=0; %迭代次數初值 alpha=2/(1+sin(pi/num));%超松弛迭代因子 k=r1/r2; %兩介質電導率之比 U(1,1:n)=up; %求解區域上邊界第一類邊界條件 U(n,1:n)=under; %求解區域下邊界第一類邊界條件 U(2:num,1)=0;U(2:num,n)=0; for i=2:num U(i,2:num)=up-(up-under)/num*(i-1);%采用線性賦值對上下邊界之間的節點賦迭代初值 end G=1; while G>0 %迭代條件：不滿足相對誤差限要求的節點數目G不為零 Un=U; %完成第n次迭代后所有節點處的值 G=0; %每完成一次迭代將不滿足相對誤差限要求的節點數目歸零 for j=1:n for i=2:num U1=U(i,j); %第n次迭代時網格節點處的值 if j==1 %第n+1次迭代左邊界第二類邊界條件 U(i,j)=1/4*(2*U(i,j+1)+U(i-1,j)+U(i+1,j)); end if (j>1)&&(j U2=1/4*(U(i,j+1)+ U(i-1,j)+ U(i,j-1)+ U(i+1,j)); U(i,j)=U1+alpha*(U2-U1); %引入超松弛迭代因子后的網格節點處的值 end if i==n+1-j %第n+1次迭代兩介質分界面（與網格對角線重合）第二類邊界條件 U(i,j)=1/4*(2/(1+k)*(U(i,j+1)+U(i+1,j))+2*k/(1+k)*(U(i-1,j)+U(i,j-1))); end if j==n %第n+1次迭代右邊界第二類邊界條件 U(i,n)=1/4*(2*U(i,j-1)+U(i-1,j)+U(i+1,j)); end end end N=N+1 %顯示迭代次數 Un1=U; %完成第n+1次迭代后所有節點處的值 err=abs((Un1-Un)./Un1);%第n+1次迭代與第n次迭代所有節點值的相對誤差 err(1,1:n)=0; %上邊界節點相對誤差置零 err(n,1:n)=0; %下邊界節點相對誤差置零 G=sum(sum(err>deta))%顯示每次迭代后不滿足相對誤差限要求的節點數目G end
標簽： 有限差分
上傳時間： 2018-07-13
上傳用戶：Kemin
