提交记录 3341
| 提交时间 |
评测时间 |
| 2018-07-13 23:11:41 |
2020-07-31 21:15:34 |
#include <stdio.h>
inline unsigned g() {
unsigned t = getchar()-48;
t=t*10+ getchar()-48;
t=t*10+ getchar()-48;
return t*10+ getchar()-48;
}unsigned long long x[8500];
int main() {
unsigned a[2500], b[2500];
int i,j;unsigned k;unsigned long long cy=0;
for(i=0;i<2500;i++)a[i]=g();getchar();
for(i=0;i<2500;i++){
k=g();
for(j=0;j<2500;j++)x[i+j+1]+=(unsigned long long)a[i]*k;
}//if(getchar()!=10)while(1)putchar('s');
for(i=5000;i--;){cy+=x[i];x[i]=cy%10000,cy/=10000;}//while(x[0]==0);
printf("%d",x[0]);for(i=1;i<5000;i++){
printf("%04d",x[i]);
}
}
| Compilation | N/A | N/A | Compile OK | Score: N/A | 显示更多 |
| Testcase #1 | 2.874 ms | 80 KB | Wrong Answer | Score: 0 | 显示更多 |
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