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;
}