ZOJ 1057 Undercut（简单模拟）

[b]Undercut[/b]

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[20],b[20];
while(scanf("%d",&n) && n){
if(!first) puts("");
for(i=0; i<n; i++)
scanf("%d", &a[i]);
for(i=0; i<n; i++)
scanf("%d", &b[i]);
int suma =0 ,sumb=0;
for(i=0; i<n; i++){
if(a[i]==1 && b[i]==2){
suma += 6;
continue;
}
if(a[i]==2 && b[i]==1){
sumb += 6;
continue;
}
if(a[i]==b[i]) continue;
if(a[i] - b[i] == 1){
sumb += (a[i]+b[i]);
continue;
}
if(b[i] - a[i] == 1){
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 0;
}