hdu-2509-反nim博弈
阿新 • • 發佈:2018-07-22
single otto contest ont let 註意 ati any pac
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.
Sample Input
2
2 2
1
3
Be the Winner
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4699 Accepted Submission(s): 2576
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).
Output If a winning strategies can be found, print a single line with "Yes", otherwise print "No".
Sample Output No Yes
Source ECJTU 2008 Autumn Contest
一開始沒註意到0態是勝態這個條件,想成sg來做了,結果WA。
這個題目參考1709的證明過程,不同的地方在於這個可以將一堆取走一部分之後再把剩下的分成兩堆,有趣啊,
但是按照上述證明即使有這個條件依然不影響結論的正確性,所以可以直接套用。
1 #include<bits/stdc++.h> 2 using namespace std; 3 int main(){ 4 intt,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