标签：clu ups cep bit one ott return arc 题目
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4699 Accepted Submission(s): 2576
Problem Description
Let‘s consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).
Input
You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.
Output
If a winning strategies can be found, print a single line with "Yes", otherwise print "No".
Sample Input
Sample Output
Source
一开始没注意到0态是胜态这个条件，想成sg来做了，结果WA。
这个题目参考1709的证明过程，不同的地方在于这个可以将一堆取走一部分之后再把剩下的分成两堆，有趣啊，
但是按照上述证明即使有这个条件依然不影响结论的正确性，所以可以直接套用。
1 #include<bits/stdc++.h>
2 using namespace std;
3 int main(){
4 int t,n,m;
5 while(cin>>n){
6 int a,sg=0,tot=0;
7 for(int i=1;i<=n;++i){
8 cin>>a;
9 sg^=a;
10 if(a==1) tot++;
11 }
12 if( (tot==n&&sg==0) || (tot!=n&&sg!=0)) puts("Yes");
13 else puts("No");
14 }
15 return 0;
16 }
hdu-2509-反nim博弈
标签：clu ups cep bit one ott return arc 题目
原文地址：https://www.cnblogs.com/zzqc/p/9349904.html
