2014 Multi-University Training Contest 5

官方解题报告http://blog.sina.com.cn/s/blog_a19ad7a10102uyxr.html

 1 #include<cstdio>
 2 #include<cctype>
 3 const int M=810;
 4 int a[M][M],b[M][M],c[M][M];
 5 inline int getint(){
 6     int res=0;
 7     char tmp;
 8     while(!isdigit(tmp=getchar()));
 9     do{
10         res=(res<<3)+(res<<1)+tmp-'0';
11     }while(isdigit(tmp=getchar()));
12     return res;
13 }
14 int main(){
15     int n,x,i,j,k;
16     while(~scanf("%d",&n)){
17         for(i=0;i<n;i++){
18             for(j=0;j<n;j++){
19                 a[i][j]=getint();
20                 a[i][j]%=3;
21             }
22         }
23         for(i=0;i<n;i++){
24             for(j=0;j<n;j++){
25                 b[i][j]=getint();
26                 b[i][j]%=3;
27                 c[i][j]=0;
28             }
29         }
30         for(k=0;k<n;k++){
31             for(i=0;i<n;i++){
32                 if(a[i][k]){
33                     for(j=0;j<n;j++){
34                         c[i][j]+=a[i][k]*b[k][j];
35                     }
36                 }
37             }
38         }
39         for(i=0;i<n;i++){
40             for(j=0;j<n;j++){
41                 if(j) putchar(' ');
42                 putchar(c[i][j]%3+'0');
43             }
44             putchar('\n');
45         }
46     }
47     return 0;
48 }

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