解题思路:
最小公倍数,匠气的代码...
注意事项:
参考代码:
#define _CRT_SECURE_NO_WARNINGS #include <iostream> #include <vector> #include <algorithm> #define N 30 using namespace std; int factors_raw[] = { 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 }; int factors[N]; vector<int> numArr; int main(int argc, char** argv) { for (int i = 0; i < sizeof(factors_raw)/sizeof(int); i++) factors[factors_raw[i]] = factors_raw[i]; int a, b, c; cin >> a >> b >> c; //不都是素数 while (factors[a] == 0 || factors[b] == 0 || factors[c] == 0) { for (int i = 2; i <= max(max(a, b),max(b, c)); i++) { bool okPrime = false; if (factors[i] != 0) { if (a%i == 0) { a /= i; okPrime = true; } if (b%i == 0) { b /= i; okPrime = true; } if (c%i == 0) { c /= i; okPrime = true; } if (okPrime) { numArr.push_back(i); } } } } int product = a*b*c; for (vector<int>::iterator it = numArr.begin(); it < numArr.end(); it++) product *= *it; cout << product << endl; return 0; }
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