PAT 天梯赛 L1-009. N个数求和 【模拟】

题目链接

https://www.patest.cn/contests/gplt/L1-009

思路

每一步每一步 往上加，但是要考虑 溢出，所以用 LONG LONG 而且 每一步 都要约分 才能保证不溢出 最后要考虑 整数部分 和分子部分都为0的情况

AC代码

#include <cstdio>

#include <cstring>

#include <ctype.h>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#include <cmath>

#include <deque>

#include <vector>

#include <queue>

#include <string>

#include <map>

#include <stack>

#include <set>

#include <numeric>

#include <sstream>

using namespace std;

typedef long long LL;

const double PI = 3.14159265358979323846264338327;

const double E = 2.718281828459;

const double eps = 1e-6;

const int MAXN = 0x3f3f3f3f;

const int MINN = 0xc0c0c0c0;

const int maxn = 1e5 + 5;

const int MOD = 1e9 + 7;

LL gcd(LL x, LL y)

{

    LL r;

    while (1)

    {

        r = x % y;

        if (r == 0)

            break;

        x = y;

        y = r;

    }

    return y;

}

int main()

{

    int n;

    cin >> n;

    LL a, b, c, d;

    scanf("%lld/%lld", &a, &b);

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

    {

        scanf("%lld/%lld", &c, &d);

        LL temp = b * d /gcd(b, d);

        a *= (temp / b);

        a += (c * (temp / d));

        b = temp;

        temp = gcd(a, b);

        a /= temp;

        b /= temp;

    }

    LL vis = a / b;

    a %= b;

    if (vis)

    {

        printf("%lld", vis);

        if (a)

        {

            if (vis < 0 && a > 0)

                a *= -1;

            printf(" %lld/%lld", a, abs(b));

        }

    }

    else if(a)

    {

        if (b < 0 && a > 0)

            a *= -1, b *= -1;

        printf("%lld/%lld", a, abs(b));

    }

    else

        cout << 0 ;

    cout << endl;

}