HDU 1847 good luck in CET-4 everybody! (Bashi game theory)

Address: HDU 1847

This problem can be converted using NP state.

First, 0 means that you cannot play the cards. Then, it is a winning state that can reach the "fail" state in one step. What cannot reach the "fail" state in one step is a "fail" state. The state transition equation can be introduced and then solved using DP. That is, the transition from a known state to an unknown state is a small transition to a large state. If its next step is not defeated, it will be defeated, that is, it must win.

The Code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;int dp[2000], a[100];int main(){    int n, i, j, m, flag;    a[0]=1;    for(i=1;; i++)    {        a[i]=a[i-1]*2;        if(a[i]>1000)        {            m=i;            break;        }    }    dp[0]=0;    for(i=1;i<=1000;i++)    {        flag=0;        for(j=0;j<m;j++)        {            if(a[j]>i)                break;            if(!dp[i-a[j]])            {                flag=1;                break;            }        }        if(flag)            dp[i]=1;        else            dp[i]=0;    }    while(scanf("%d",&n)!=EOF)    {        if(dp[n])            puts("Kiki");        else            puts("Cici");    }    return 0;}

HDU 1847 good luck in CET-4 everybody! (Bashi game theory)