C语言编程重点1

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1、1,.求【mn】中既不能被7整除也不能被5整除的和#include"stdio.h"intfun(intm,intn){/***********begin***********/inti,sum=0;for(i=m;i<=n;i++){if(((i%7)!=0)&&((i%5)!=0))sum=sum+i;}returnsum;/************end************/voidmain(){voidNONO();//函数声明ints;intm,n;printf("Entermn:");scanf("%d%d",&m,&n);s=fun(m,n);printf("s=

2、%d",s);NONO();}voidNONO(){FILE*fr,*fw;inti,x,y;fr=fopen("D:\exam\03910431\PROGIN.DAT","r");fw=fopen("D:\exam\03910431\PROGOUT.DAT","w");for(i=1;i<=5;i++){fscanf(fr,"%d,%d",&x,&y);fprintf(fw,"TheResultis:%d***",fun(x,y));}fclose(fr);fclose(fw);}2.求3*3矩阵的最小值#include   #define  N

3、  3  int fun(int a[N][N])  {  int min;  /***********begin***********/int i,j;min=a[0][0];for(i=0;i

4、 printf("array is:");    for(i=0;i

5、   fw=fopen("D: \exam\03910411\PROGOUT.DAT","w");    for(i=1;i<=5;i++)    { for(j=0;j

6、doublefun(intn){doublec;/***********begin***********/c=0.0;inti,t=1;for(i=1;i<=n;i++){c=c+t*1.0/(2*i-1);t=-t;}returnc;/***********end***********/}voidmain(){voidNONO();//函数声明doublesum;intn;printf("Entern:");scanf("%d",&n);sum=fun(n);printf("sum=%.4f",sum);NONO();}voidNONO(){FILE*fr,*fw;inti,

7、x;fr=fopen("D:\exam\03910304\PROGIN.DAT","r");fw=fopen("D:\exam\03910304\PROGOUT.DAT","w");for(i=1;i<=5;i++){fscanf(fr,"%d",&x);fprintf(fw,"s=%.4f",fun(x));}fclose(fr);fclose(fw);}4.将数组a的5个元素对折后再输出，比如：a[5]={1,2,3,4,5