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;
}