Codeforces 389A Fox and Number Game(贪心)

题意:给出一个序列X,能进行如下操作,选择两个下i, j(i != j), 将Xi的值减去Xj, 求怎么样操作能使的最后序列和最小.

思路:做法是每次都找序列里最大的和次大的,最大的减掉次大的,如果都变成了相同的大小,就退出循环.

#include <cstdio>#include <algorithm>#include <numeric>using namespace std;const int MAX = 101;int X[MAX];int main(int argc, char const *argv[]){int n, ans = 0;scanf("%d", &n);for(int i = 0; i < n; ++i){scanf("%d", &X[i]);}while(true){int mp = max_element(X, X + n) - X;int smp = mp, smv = 0;for(int i = 0; i < n; ++i){if(X[i] < X[mp] && smv < X[i]){smv = X[i];smp = i;}}if(mp == smp)break;X[mp] -= X[smp];}printf("%d\n", accumulate(X, X + n, 0));return 0;}