hdu 1099通分

这题几年前就看过，一直没看懂啊。。。今天翻解题报告，才知道就是求n/1、n/2、n/3、...、n/(n-1)、n/n的和，并且以分数的形式表示，还算简单。

/*

 * hdu1099/win.cpp

 * Created on: 2012-7-27

 * Author    : ben

 */

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <queue>

#include <set>

#include <map>

#include <stack>

#include <string>

#include <vector>

#include <deque>

#include <list>

#include <functional>

#include <numeric>

#include <cctype>

using namespace std;

int gcd(int a, int b) {

    int r;

    while(b) {

        r = a % b;

        a = b, b = r;

    }

    return a;

}



int getbitlen(int n) {

    char str[300];

    sprintf(str, "%d", n);

    return strlen(str);

}



int main() {

#ifndef ONLINE_JUDGE

    freopen("data.in", "r", stdin);

#endif

    int n, up, down, intpart;

    while(scanf("%d", &n) == 1) {

        intpart = up = 0;

        down = 1;

        for(int i = 1; i <= n; i++) {

            up = up * i + n * down;

            down *= i;

            int temp = gcd(up, down);

            up /= temp;

            down /= temp;

            temp = up / down;

            up -= temp * down;

            intpart += temp;

        }

        if(up == 0) {

            printf("%d\n", intpart);

        }else {

            int len = getbitlen(intpart);

            for(int i = 0; i <= len; i++) {

                putchar(' ');

            }

            printf("%d\n", up);

            printf("%d ", intpart);

            len = getbitlen(up) > getbitlen(down) ? getbitlen(up) : getbitlen(down);

            for(int i = 0; i < len; i++) {

                putchar('-');

            }

            putchar('\n');

            len = getbitlen(intpart);

            for(int i = 0; i <= len; i++) {

                putchar(' ');

            }

            printf("%d\n", (int)down);

        }

    }

    return 0;

}