取(m堆)石子游戏
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3598 Accepted Submission(s): 2151
Problem Description
m堆石子,两人轮流取.只能在1堆中取.取完者胜.先取者负输出No.先取者胜输出Yes,然后输出怎样取子.例如5堆 5,7,8,9,10先取者胜,先取者第1次取时可以从有8个的那一堆取走7个剩下1个,也可以从有9个的中那一堆取走9个剩下0个,也可以从有10个的中那一堆取走7个剩下3个.
Input
输入有多组.每组第1行是m,m<=200000. 后面m个非零正整数.m=0退出.
Output
先取者负输出No.先取者胜输出Yes,然后输出先取者第1次取子的所有方法.如果从有a个石子的堆中取若干个后剩下b个后会胜就输出a b.参看Sample Output.
Sample Input
2
45 45
3
3 6 9
5
5 7 8 9 10
0
Sample Output
No Yes 9 5 Yes 8 1 9 0 10 3
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
", a[p1[i]], a[p1[i]] - p2[i]);
}
}
}
return 0;
}