Combo Game Simple problem

Uva-11859-division Game

Topic Delivery: Division Game

AC Code:

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL Long Long#define INF 0x7fffffffusing namespace STD;Const intMAXN =10005;intN, M, T;intmatch[ -];//Put a number of rows into their true factor, that is, take away some of the vegetarian factor, that can be seen as a bunch of matches to take out a few, so it can be seen as a combination of Nim gameintCHU[MAXN];intTO[MAXN];//Calculate the number of elements per numbervoidInit () {//Hit the table to calculate the element factorto[1] =0; for(inti =2; i < MAXN; i + +) chu[i] = i; for(inti =2; i < MAXN; i + +) {if(To[i])Continue; To[i] + +; for(intj =2I J < Maxn; J + = i) { while(Chu[j]% i = =0) {To[j] + +;            CHU[J]/= i; }        }    }}intMain () {init ();intTscanf("%d", &t); for(intCAS =1; CAS <= T; CAS + +) {memset(Match,0,sizeof(match));scanf("%d%d", &n, &m); for(inti =1; I <= N; i + +) { for(intj =1; J <= M; J + +) {scanf("%d", &t);            Match[i] + = to[t]; }        }//Because the title can be considered as Nim game, so the SG value is the match value of the XOR and        intSG = match[1]; for(inti =2; I <= N;        i + +) {sg = sg ^ match[i]; }if(sg = =0)printf("Case #%d:no\n", CAs);//sg is 0 o'clock, the winner will be defeated, and vice versa.        Else printf("Case #%d:yes\n", CAs); }return 0;}

Uvalive-5059-playing with stones

Topic Transmission: Playing with stones

This problem can be a table to find the law.

To play the table code:

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL Long Long#define INF 0x7fffffffusing namespace STD;Const intMAXN = -;intSG[MAXN];intVIS[MAXN];intMain () {sg[1] =0;//Only one time can not be taken, for the state of defeat     for(inti =2; I <= -; i + +) {memset(Vis,0,sizeof(VIS)); for(intj =1; J *2<= i; J + +) Vis[sg[i-j]] =1; for(intj =0;; J + +)if(!vis[j]) {Sg[i] = j; Break; }printf("%d", Sg[i]); }return 0;}

The law is:
When n is even, SG (n) = N/2;
When n is odd, sg (n) = SG (N/2);

AC Code:

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL Long Long#define INF 0x7fffffffusing namespace STD; ll SG (ll x) {return(X &1) ? SG (X/2): X/2;}intMain () {intTCin>> T; while(T-) {intN LL A, ans =0;Cin>> N; for(inti =0; I < n; i + +) {Cin>> A;        Ans ^= SG (a); }if(ANS)cout<<"yes\n";Else cout<<"no\n"; }return 0;}

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Combo Game Simple problem