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：

 # include<stdio.h>
 # include<string.h>
 int main(){
     bool first = true;
     int n,i;
     int a[],b[];
     while(scanf("%d",&n) && n){
         if(!first) puts("");
         for(i=; i<n; i++)
             scanf("%d", &a[i]);
         for(i=; i<n; i++)
             scanf("%d", &b[i]);
         int suma = ,sumb=;
         for(i=; i<n; i++){
             if(a[i]== && b[i]==){
                 suma += ;
                 continue;
             }
             if(a[i]== && b[i]==){
                 sumb += ;
                 continue;
             }
             if(a[i]==b[i]) continue;
             if(a[i] - b[i] == ){
                 sumb += (a[i]+b[i]);
                 continue;
             }
             if(b[i] - a[i] == ){
                 suma += (a[i]+b[i]);
                 continue;
             }
             if(a[i]>b[i]) suma += a[i];
             if(b[i]>a[i]) sumb += b[i];
         }
         printf("A has %d points. B has %d points.\n",suma,sumb);
         first = false;
     }
     return ;
 }