nyoj1013 除法表达式

题目：点击打开链接

 

分析：可易知，X1一定在分子上，X2一定在分母上，要想使Ｅ的结果为整数，即Xk尽量要在分子的上，可将Ｅ改写成：X1/X2/(X3/X4/X5/X6.../Xk)=(X1*X3*X4*X5*X6...*Xk)/X2，现在简单了吧？

 

方法１：用大数相乘，再用大数求余

方法２：从Ｘ１到Ｘｋ依次跟Ｘ２约分，知道Ｘ２为１时结束．

 

上码：

#include<iostream>#include<cstdio>#include<vector>#include<algorithm> int gcd(int a, int b) {return b ? gcd(b, a % b) : a;} using namespace std; int main() {int nCase, n, x2;char c;bool is_integer;vector<int> myVector;cin >> nCase; while (nCase--) {myVector.clear(); while (1) {scanf("%d%c",&n,&c);myVector.push_back(n); if (c == '\n') {break;}} if (myVector.size() == 1) {is_integer = true;} else {x2 = myVector[1]; for (vector<int>::size_type i = 0; x2 != 1 && i < myVector.size(); i++) {if (i == 1) {continue;} x2 /= gcd(x2, myVector[i]);} x2 == 1 ? is_integer = true : is_integer = false;} is_integer ? cout << "YES" << endl : cout << "NO" << endl;} return 0;}