# 51nod Game Theory Water problem

Source: Internet
Author: User

51nod1069 Nim Games

There are n heaps of stones. A B Two people take turns, a take first. Each time can only from a bunch of a few, can be a bunch of all take away, but not to take, the last 1 stones to win the people. Suppose a B is very clever, there is no mistake in the process of taking the stone. Give the number of N and each heap of stones, and ask who can win the game at the end. For example: 3 piles of gravel, 1 capsules per heap. A take 1, B take 1, at this time there are 1 piles left, so a can get the last 1 stones.
`/* For a situation (A1,A2,..., an), if a1^a2^...^an<>0, there must be some legal movement, the AI will be changed into AI ' after A1^a2^...^ai ' ^...^an=0. As a a1^a2^...^an=k, there must be an AI, whose binary representation is 1 at the top of the K (otherwise the highest bit of k is how the 1 is obtained). Then Ai^k<ai must be set up. Then we can change AI to Ai ' =ai^k, at this time A1^a2^...^ai ' ^...^an=a1^a2^...^an^k=0. *///O (n) to find a winning strategy. #include <cstdio> #include <cstring> #include <cctype> #include <algorithm>using namespace std; #define REP (i,s,t) for (int i=s;i<=t;i++) int read () {int X=0;char C=getchar (), while (!isdigit (c)) C=getchar (), while ( IsDigit (c)) x=x*10+c-' 0 ', C=getchar (); return x;} int main () {int n=read (), T,ans=0;rep (i,1,n) ans^= (T=read ()), if (ans) puts ("A"), Else puts ("B"); return 0;}`

51nod1067 Bash Game V2

There are a bunch of stones in total N. A B Two people take turns, a take first. Each time can only take 1,3,4, to get the last 1 stones of the people to win. Suppose a B is very clever, there is no mistake in the process of taking the stone. Give N and ask who can win the game at the end. For example n = 2. A can only take 1, so B can get the last 1 stones.
`101111010111101011110#include<cstdio> #define REP (i,s,t) for (int i=s;i<=t;i++) int main () {int n,u;scanf ("%") D ", &n); Rep (i,1,n) {scanf ("%d ", &u); U%=7;if (u==2| | u==0) puts ("B"); else puts ("A");} return 0;}`

51nod1185 Witzov Game V2

There are 2 piles of stones. A B Two people take turns, a take first. Each time you can take any or 2 of the same number of stones from a heap, but not to be taken. The man who got the last 1 stones won. Suppose a B is very clever, there is no mistake in the process of taking the stone. Give the number of 2 stones and ask who will win the game at the end. For example, 2 stones were 3 and 5. So no matter a how to take, B has a corresponding method to get the last 1.
`It is related to the Golden Section to find the law of the table. This card accuracy needs to be simulated multiplication ... Good God's analog multiplication ... #include <cstdio> #include <cstring> #include <cctype> #include <algorithm> #include <cmath >using namespace std; #define REP (I,s,t) for (int. i=s;i<=t;i++) #define DWN (i,s,t) for (int i=s;i>=t;i--) #define ll long Longll Read () {ll x=0;char C=getchar (); while (!isdigit (c)) C=getchar (); while (IsDigit (c)) x=x*10+c-' 0 ', C=getchar (); return x;} LL Ans={618033988,749894848,204586834};const ll Mod=1e9;int main () {int t=read (); LL a,b,aa,ab,ac,ad,ba,bb;double TP = (sqrt (5) +1)/2;while (t--) {a=read (), B=read (); if (a>b) swap (A, b); Ba= (b-a)/mod,bb= (b-a)%mod;aa=bb*ans;ab=ba* Ans+aa/mod+bb*ans;ac=ba*ans+ab/mod+bb*ans;ad=b-a+ba*ans+ac/mod;if (Ad==a) puts ("B"); else puts ("a") ;} return 0;}`

51nod Game Theory Water problem

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