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; }转载于:https://www.cnblogs.com/Dup4/p/9433268.html
