ZOJ 1057 Undercut（简单模拟）

Undercut

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=57

题目大意：a card game,two player,每个player有n张牌，按顺序出牌，如果2张牌的点数相等，2个player都不得分；如果2张牌点数相差1，点数大的player得分为2张牌的点数和；如果2张牌的点数相差超过1，点数大的player得分为点数大的牌的点数；特殊情况，如果2张牌的点数分别为1和2，持2的player得分为6而不是3；给出2个player各n张牌的点数，要求输出每个player的得分情况

代码如下：

 1 # include<stdio.h>
 2 # include<string.h>
 3 int main(){
 4     bool first = true;
 5     int n,i;
 6     int a[20],b[20];
 7     while(scanf("%d",&n) && n){
 8         if(!first) puts("");
 9         for(i=0; i<n; i++)
10             scanf("%d", &a[i]);
11         for(i=0; i<n; i++)
12             scanf("%d", &b[i]);
13         int suma =0 ,sumb=0;
14         for(i=0; i<n; i++){
15             if(a[i]==1 && b[i]==2){
16                 suma += 6;
17                 continue;
18             }
19             if(a[i]==2 && b[i]==1){
20                 sumb += 6;
21                 continue;
22             }
23             if(a[i]==b[i]) continue;
24             if(a[i] - b[i] == 1){
25                 sumb += (a[i]+b[i]);
26                 continue;
27             }
28             if(b[i] - a[i] == 1){
29                 suma += (a[i]+b[i]);
30                 continue;
31             }
32             if(a[i]>b[i]) suma += a[i];
33             if(b[i]>a[i]) sumb += b[i];
34         }
35         printf("A has %d points. B has %d points.
",suma,sumb);
36         first = false;
37     }
38     return 0;
39 }