CodeForces 342A Xenia and Divisors

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题目挺水，只有“1 2 4”，”1 2 6“，”1 3 6“这三种情况符合。所以就简单多了。

#include<stdio.h>#include<iostream>#include<math.h>#include<string.h>#include<iomanip>#include<stdlib.h>#include<ctype.h>#include<algorithm>#include<deque>#include<functional>#include<iterator>#include<vector>#include<list>#include<map>#include<queue>#include<set>#include<stack>#define CPY(A, B) memcpy(A, B, sizeof(A))typedef long long LL;typedef unsigned long long uLL;const int MOD = int (1e9) + 7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const double EPS = 1e-9;const double OO = 1e20;const double PI = acos (-1.0);const int dx[] = {-1, 0, 1, 0};const int dy[] = {0, 1, 0, -1};using namespace std;int main() {    int n,k,a[7]= {0};    while (cin>>n) {        bool f=true;        for (int i=0; i<n; i++) {            scanf ("%d",&k);            if (k==5||k==7) {f=false;}//5，7，直接排除            a[k]++;        }        if (f&&a[1]==a[2]+a[3]&&a[1]==a[4]+a[6]&&a[2]>=a[4]) {            for (int i=0; i<a[4]; i++) {                printf ("1 2 4\n");                a[2]--;a[1]--;//减去这部分2，剩余2的数量就是”1 2 6“的数量            }            for (int j=0; j<a[2]; j++) {                printf ("1 2 6\n");                a[1]--;            }            for (int k=0; k<a[1]; k++) {                printf ("1 3 6\n");            }        } else {cout<<"-1\n";}    }    return 0;}

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