【UVA】10288-Coupons（概率）

做数学简直无爱，用sprintf将数字转化成字符串计算其长度。

另外求化简2个数，直接求其最大公约数，除就行了。

打表过就行了。

14032844

10288

Coupons

Accepted

C++

0.016

2014-08-13 10:59:00

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<cmath>
#include<string>
#include<sstream>
#include<ctime>
using namespace std;
#define _PI acos(-1.0)
#define INF (1 << 10)
#define esp 1e-6
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pill;
/*===========================================
===========================================*/
#define MAXD 40
LL X[MAXD],Y[MAXD],Z[MAXD];
LL gcd(LL a, LL b){
    return  b == 0 ? a : gcd(b , a % b);
}
void _init(){
    X[1] = 1 ; Y[1] = 1;
    for(LL i = 2 ; i <= 33 ; i ++){
        Y[i] = Y[i - 1] * i;
        X[i] = X[i - 1] * i + Y[i - 1];
        LL mod = gcd(X[i],Y[i]);
        X[i] /= mod;
        Y[i] /= mod;
    }
    for(int i = 1 ; i <= 33 ; i ++){
        X[i] = X[i] * i;
        Z[i] = X[i] / Y[i];
        X[i] = X[i] %  Y[i];
        LL mod = gcd(X[i],Y[i]);
        X[i] /= mod;
        Y[i] /= mod;
    }
    return ;
}
int main(){
    int T,n;
    _init();
    while(scanf("%d",&n) != EOF){
        if(Z[n] > 0){
            char num1[MAXD],L1;
            sprintf(num1,"%lld",Z[n]);
            L1 = strlen(num1);
            if(X[n] > 0){
                sprintf(num1,"%lld",Y[n]);
                int L2 = strlen(num1);
                for(int i = 0 ; i <= L1 ; i++)
                    printf(" ");
                printf("%lld\n",X[n]);
                printf("%lld ",Z[n]);
                for(int i = 0 ; i < L2 ; i ++)
                    printf("-");
                printf("\n");
                for(int i = 0 ; i <= L1 ; i++)
                    printf(" ");
                printf("%lld\n",Y[n]);
            }
            else{
                printf("%lld\n",Z[n]);
            }
        }
        else{
            char num1[MAXD],L1;
            sprintf(num1,"%lld",Z[n]);
            L1 = strlen(num1);
            if(X[n] > 0){
                sprintf(num1,"%lld",Y[n]);
                int L2 = strlen(num1);
                for(int i = 0 ; i <= L1 ; i++)
                    printf(" ");
                printf("%lld\n",X[n]);
                printf("%lld ",Z[n]);
                for(int i = 0 ; i < L2 ; i ++)
                    printf("-");
                printf("\n");
                for(int i = 0 ; i <= L1 ; i++)
                    printf(" ");
                printf("%lld\n",Y[n]);
            }
        }
    }
    return 0;
}