117的5#include<stdio.h>#defineN100intflag[N];intcount[N];voidfindmax(int*num,intn){inttemp,i,j;for(i=0;i<n;i++){temp=*(num+i);if(flag[i]==0){for(j=i;j<n;j++)if(*(num+j)==temp){flag[j]=1;count[i]++;}}}}voidmain(){inta[N];intnmax=0;intn,i;printf("inputthesizeofthearray:");scanf("%d",&n);for(i=0;i<n;i++){printf("第%d个:",i+1);scanf("%d",&a[i]);flag[i]=0;count[i]=0;}findmax(a,n);nmax=count[0];for(i=0;i<n;i++)if(count[i]>nmax)nmax=count[i];for(i=0;i<n;i++)if(count[i]==nmax)printf("%d%d\n",a[i],count[i]);}105的6P151输入两个数组（数组元素个数自定），输出在两个数组中都出现的元素。#include"stdio.h"#defineNA6#defineNB8voidmain(){floata[NA],b[NB];inti,j;for(i=0;i<NA;i++)scanf("%f",&a[i]);for(i=0;i<NB;i++)scanf("%f",&b[i]);for(i=0;i<NA;i++)for(j=0;j<NB;j++)if(a[i]==b[j]){printf("%f\n",a[i]);break;}编程，将字符数组S2中的全部字符拷贝到字符数组S1中。#include"stdio.h"voidmain(){chars1[20],s2[]="Goodmorning!";inti=0;while((s1[i++]=s2[i])!='\0');printf("%s\n",s1);}写一个函数，统计m行n列二维数组中有多少个正数、多少个负数，多少个零，并返回统计结果。#include<stdio.h>voidsub(float**a,intm,intn,int*fs,int*lin,int*zs){inti,j;*fs=*lin=*zs=0;for(i=0;i<m;i++)for(j=0;j<n;j++)if(*(*(a+i)+j)<0)(*fs)++;elseif(*(*(a+i)+j)==0)(*lin)++;else(*zs)++;return;}voidmain()//函数引用示例{floatb[5][3]={{-1,5,2},{3,0,-2},{0,-3,5},{4,7,-8},{3,4,5}},*c[5];inti,k1,k2,k3;for(i=0;i<5;i++)c[i]=b[i];sub(c,5,3,&k1,&k2,&k3);printf("负数%d个，零%d个，正数%d个。\n",k1,k2,k3);}1．编程，统计在所输入的50个实数中有多少个正数、多少个负数、多少个零。#include"stdio.h"#defineN50voidmain(){floatx;unsignedints1,s2,s3,i;s1=s2=s3=0;for(i=1;i<=N;i++){scanf("%f",&x);if(x<0)s1++;elseif(x==0)s2++;elses3++;}printf("负数%u个，零%u个，正数%u个\n",s1,s2,s3);}2.编程，计算并输出方程X2+Y2=1989的所有整数解。#include"stdio.h"voidmain(){intx,y;for(x=-45;x<=45;x++){y=-45;while(y<=45){if(x*x+y*y==1989)printf("%d*%d+%d*%d=%d\n",x,x,y,y,1989);y++;}}}3．编程，输入一个10进制正整数，然后输出它所对应的八进制、十六进制数。#include"stdio.h"voidmain(){unsignedintx;printf("请输入一个十进制正整数:");scanf("%u",&x);printf("%d=八进制数%o=十六进制数%x\n",x,x,x);}4．编程，找出1000以内的所有完数，并输出其因子。#include"stdio.h