[C](0.208)
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| #include<stdio.h> | |
| #include<string.h> | |
| int win( char *m1, char *m2 ) | |
| { | |
| if( (m1[0] == 'r' && m2[0] == 's') || (m1[0] == 's' && m2[0] == 'p') || (m1[0] == 'p' && m2[0] == 'r') ) | |
| return 1; | |
| return 0; | |
| } | |
| int main() | |
| { | |
| int n, k; | |
| int print = 0; | |
| while( scanf( "%d", &n ) != EOF && n != 0 ) | |
| { | |
| scanf( "%d", &k ); | |
| if( print ) | |
| printf( "\n" ); | |
| print = 1; | |
| int people[105][2] = {0}; | |
| int p1, p2; | |
| char m1[10], m2[10]; | |
| int i; | |
| for( i = 0 ; i < k*n*(n-1)/2 ; i++ ) | |
| { | |
| scanf( "%d%s%d%s", &p1, m1, &p2, m2 ); | |
| if( win( m1, m2 ) ) | |
| { | |
| people[p1][0]++; | |
| people[p2][1]++; | |
| } | |
| else if( win( m2, m1 ) ) | |
| { | |
| people[p1][1]++; | |
| people[p2][0]++; | |
| } | |
| } | |
| for( i = 1 ; i <= n ; i++ ) | |
| if( people[i][0]+people[i][1] != 0 ) | |
| printf( "%.3lf\n", (double)(people[i][0])/(double)(people[i][0]+people[i][1]) ); | |
| else | |
| printf( "-\n" ); | |
| } | |
| return 0; | |
| } |
1 意見:
可以跟我說一下這一題的題意是甚麼嗎?我看了很久都看不懂
張貼留言