1:编程对用户输入一个含有10个整数的一维数组,分别计算数组中所有正数和所有负数的和。 编程对用户输入一个含有10个整数的一维数组,分别计算数组中所...

\u7528c++\u7f16\u7a0b\u5bf9\u7528\u6237\u8f93\u5165\u4e00\u4e2a\u542b\u670910\u4e2a\u6574\u6570\u7684\u4e00\u5806\u6570\u7ec4\uff0c\u5206\u522b\u8ba1\u7b97\u6570\u7ec4\u4e2d\u6240\u6709\u6b63\u6570\u548c\u6240\u6709\u8d1f\u6570\u7684\u548c \u6c42\u5927\u795e\u544a

#include using namespace std;int main() { int num[10] = {0}; int Zheng = 0, Fu = 0; for (int i = 0; i > num[i]; if (num[i] > 0) Zheng += num[i]; else Fu += num[i]; } cout << "Zheng : " << Zheng << "\tFu : " << Fu << endl; return 0;}

void main(){
int a[10];
int np=0,nn=0;
for (int i=0;i<10;i++){
scanf("%d",&a[i]);
if (a[i]>0) np+=a[i];
else if (a[i]<0) nn+=a[i];
}
printf("the sum of positives is %d, the sum of negatives is %d\n",np,nn);
}

2
#include "stdio.h"
#include "string.h"
void main(){
char str[3][21],*p;
for (int i=0;i<3;i++){
gets(str[i]);
if (i==0) p=str[0];
else if (strcmp(str[i],p)>0) p=str[i];
}
printf("%s",p);
}

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