标签:博弈 inf 整数 out else names scan cep tle
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3598 Accepted Submission(s): 2151
Nim博弈并要求给出策略
新开两个数组记录完事
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w",stdout);
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int, int> PII;
using namespace std;
int a[200005];
int p1[200005];
int p2[200005];
int main() {
//FIN
int n;
while(~scanf("%d", &n) && n) {
int ans = 0;
int cnt = 0;
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
ans ^= a[i];
}
for(int i = 1; i <= n; i++) {
if((ans ^ a[i]) < a[i]) {
p1[cnt] = i;
p2[cnt] = a[i] - (ans ^ a[i]);
cnt++;
}
}
if(ans == 0) puts("No");
else {
puts("Yes");
for(int i = 0; i < cnt; i++) {
printf("%d %d\n", a[p1[i]], a[p1[i]] - p2[i]);
}
}
}
return 0;
}
标签:博弈 inf 整数 out else names scan cep tle
原文地址:http://www.cnblogs.com/Hyouka/p/7425334.html