UVA 1636 Head Shot 概率 数学

  题目链接: https://vjudge.net/problem/UVA-1636

  题目描述: 一个左轮手枪有空弹还有实弹, 有两种操作, 直接抠下一发或者随机转一下, 给出你左轮手枪的子弹序列, 让你求一直第一发是空弹的情况下, 下一发哪种操作时空弹的可能性大

  解题思路: 很基础的概率题, 只要分别求两个的概率即可

  代码: 

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iterator>
#include <cmath>
#include <algorithm>
#include <stack>
#include <deque>
#include <map>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,-0x3f,sizeof(a))
#define fi(n) for(i=0;i<n;i++)
#define fj(m) for(j=0;j<m;j++)
#define sca(x) scanf("%d",&x)
#define ssca(x) scanf("%s",x)
#define scalld(x) scanf("%I64d",&x)
#define print(x) printf("%d\n", x)
#define printlld(x) printf("%I64d\n",x)
#define de printf("=======\n")
#define yes printf("YES\n")
#define no printf("NO\n")
typedef long long ll;
using namespace std;

const int eps = 1e-9;
const int maxn = 150;
char a[maxn];

int main() {
    while( ssca(a) != EOF ) {
        int len = (int)strlen(a);
        int n = 0;
        for( int i = 0; i < len; i++ ) {
            if( a[i] == '0' ) n++;
        }
        double ran = double(n)/double(len);
        int sum = 0;
        for( int i = 0; i < len; i++ ) {
            if( a[i] == '0' ) {
                if( a[(i+1)%len] == '0' ) {
                    sum++;
                }
            }
        }
        double sho = double(sum)/double(n);
        if( ran - sho > eps ) {
            printf( "ROTATE\n" );
        }
        else if( sho > ran > eps ) {
            printf( "SHOOT\n" );
        }
        else {
            printf( "EQUAL\n" );
        }
    }
    return 0;
}
View Code

  思考: 对条件概率的理解要加深

}

 

posted on 2017-08-23 10:57  FriskyPuppy  阅读(249)  评论(0编辑  收藏  举报

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