Codeforces Round #399 E Game of Stones 博弈
2017-11-06 21:22
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http://codeforces.com/contest/768/problem/E
题意:
给n堆石子,在一堆石头上取的石头数量只能去一次,比如你在一堆石子上取了了4个石子,那么接下里你就不能在这堆石子上取4个石子了.问后手是否必胜
做法:
╭(╯^╰)╮ 肯定是在NIM博弈基础上变了变,看了看题解。
NIM博弈就是异或和,最后看ans是否为0。对于这个题,考虑到nim中石子数量其实就是最多能取石子的步数,最后的亦或和也就是最大步数的亦或和,我们可以去求下这个游戏中每堆石子最多能取的步数,这样就能转换为普通的nim了.每堆石子最多能取的步数就好好求了,每堆石子我们按1,2,3,4..n的顺序去取石子,当剩下的石子小于等于n的时候,就不能再取了,所以n就是最大步数,然后我们去求下亦或和就可以。
题意:
给n堆石子,在一堆石头上取的石头数量只能去一次,比如你在一堆石子上取了了4个石子,那么接下里你就不能在这堆石子上取4个石子了.问后手是否必胜
做法:
╭(╯^╰)╮ 肯定是在NIM博弈基础上变了变,看了看题解。
NIM博弈就是异或和,最后看ans是否为0。对于这个题,考虑到nim中石子数量其实就是最多能取石子的步数,最后的亦或和也就是最大步数的亦或和,我们可以去求下这个游戏中每堆石子最多能取的步数,这样就能转换为普通的nim了.每堆石子最多能取的步数就好好求了,每堆石子我们按1,2,3,4..n的顺序去取石子,当剩下的石子小于等于n的时候,就不能再取了,所以n就是最大步数,然后我们去求下亦或和就可以。
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-. /// __.' ~. .~ `.__ /// .'// \./ \\`. /// .'// | \\`. /// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`. /// .'//.-" `-. | .-' "-.\\`. /// .'//______.============-.. \ | / ..-============.______\\`. /// .'______________________________\|/______________________________`. #pragma comment(linker, "/STACK:1024000000,1024000000") #include <vector> #include <iostream> #include <string> #include <map> #include <stack> #include <cstring> #include <queue> #include <list> #include <stdio.h> #include <set> #include <algorithm> #include <cstdlib> #include <cmath> #include <iomanip> #include <cctype> #include <sstream> #include <functional> #include <stdlib.h> #include <time.h> #include <bitset> using namespace std; #define pi acos(-1) #define s_1(x) scanf("%d",&x) #define s_2(x,y) scanf("%d%d",&x,&y) #define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X) #define S_1(x) scan_d(x) #define S_2(x,y) scan_d(x),scan_d(y) #define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z) #define PI acos(-1) #define endl '\n' #define srand() srand(time(0)); #define me(x,y) memset(x,y,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); cin.tie(0); #define FOR(x,n,i) for(int i=x;i<=n;i++) #define FOr(x,n,i) for(int i=x;i<n;i++) #define fOR(n,x,i) for(int i=n;i>=x;i--) #define fOr(n,x,i) for(int i=n;i>x;i--) #define W while #define sgn(x) ((x) < 0 ? -1 : (x) > 0) #define bug printf("***********\n"); #define db double #define ll long long #define mp make_pair #define pb push_back typedef long long LL; typedef pair <int, int> ii; const int INF=0x3f3f3f3f; const LL LINF=0x3f3f3f3f3f3f3f3fLL; const int dx[]={-1,0,1,0,1,-1,-1,1}; const int dy[]={0,1,0,-1,-1,1,-1,1}; const int maxn=1e3+10; const int maxx=1e6+10; const double EPS=1e-8; const double eps=1e-8; const int mod=1e9+7; template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} template <class T> inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;} while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;} inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false; while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;} else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}} if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}} if(IsN) num=-num;return true;} void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');} void print(LL a){ Out(a),puts("");} //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); //cerr << "run time is " << clock() << endl; int n; int a[maxx]; void solve() { W(cin>>n) { LL ans=0; FOR(1,n,i) { int x; s_1(x); for(int j=0;;j++) { x-=j; if(x<=j) { a[i]=j; break; } } } FOR(1,n,i) ans^=a[i]; if(ans) puts("NO"); else puts("YES"); } } int main() { //freopen( "1.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); int t=1; //init(); //s_1(t); for(int cas=1;cas<=t;cas++) { //printf("Case #%d: ",cas); solve(); } }
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