提交记录 3369
| 提交时间 |
评测时间 |
| 2018-07-14 13:34:34 |
2020-07-31 21:15:55 |
#include<stdio.h>
#include<string.h>
#define N 1010
int main(void) {
char num1[N],num2[N];
int length1,length2,c,k,i;
scanf("%s%s",num1,num2);
int a[N] = {0},b[N] = {0};
length1 = strlen(num1);
length2 = strlen(num2);
if(length1 < length2) {
k = length2;
}else {
k = length1;
}
c = k;
for(i=0; i<length1; k--,i++) {
a[k] = num1[length1 - 1 - i] - '0';//只进行到a[1],留出a[0]用以进位
}
for(i=0,k=c; i<length2; k--,i++) {
b[k] = num2[length2 - 1 - i] - '0';
}
for(i=c; i>0; i--) {
a[i] += b[i];
if(a[i]>=10) {
a[i] -= 10;
a[i-1]++;
}
}
if(a[0] != 0) {
for(i=0; i<=c; i++) {
printf("%d",a[i]);
}
}else {
for(i=1; i<=c; i++) {
printf("%d",a[i]);
}
}
printf("\n");
return 0;
}
| Compilation | N/A | N/A | Compile OK | Score: N/A | 显示更多 |
| Testcase #1 | 13.38 us | 24 KB | Wrong Answer | Score: 0 | 显示更多 |
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