把数组c和数组d对应元素相加，显示相加后的结果，如果没有对应元素，直接显示该元素本身
int c[]={3,5,8,16,16,16,22,23,37,37,50,55,67,79,85,92,98};
int d[]={93,85,78,72,72,62,53,37,30,25,17,9,8};
int n=sizeof(a)/sizeof(int);
第1个回答 2019-07-03
#include <stdio.h>
int main() {
int c[]= {3 ,5 ,8 ,16,16,16,22,23,37,37,50,55,67,79,85,92,98};
int d[]= {93,85,78,72,72,62,53,37,30,25,17,9 ,8};
int cl=sizeof(c)/sizeof(int);
int dl=sizeof(d)/sizeof(int);
printf("len-c=%d len-d=%d\n",cl,dl);
int pos=0;
int temp;
while(pos<cl || pos<dl) {
if(pos<cl && pos<dl) temp=c[pos]+d[pos];
if(pos<cl && pos>=dl) temp=c[pos];
if(pos>=cl && pos<dl) temp=d[pos];
printf("%d ",temp);
pos++;
}
}本回答被提问者和网友采纳
int main() {
int c[]= {3 ,5 ,8 ,16,16,16,22,23,37,37,50,55,67,79,85,92,98};
int d[]= {93,85,78,72,72,62,53,37,30,25,17,9 ,8};
int cl=sizeof(c)/sizeof(int);
int dl=sizeof(d)/sizeof(int);
printf("len-c=%d len-d=%d\n",cl,dl);
int pos=0;
int temp;
while(pos<cl || pos<dl) {
if(pos<cl && pos<dl) temp=c[pos]+d[pos];
if(pos<cl && pos>=dl) temp=c[pos];
if(pos>=cl && pos<dl) temp=d[pos];
printf("%d ",temp);
pos++;
}
}本回答被提问者和网友采纳
