标签：cat lines MIXED cli 3.5 nal ram eof and
zk has n numbers a1,a2,...,an. For each (i,j) satisfying 1≤i<j≤n, zk generates a new number (ai+aj). These new numbers could make up a new sequence b1，b2,...,bn(n−1)/2
6 2 2 2 4 4 4 21 1 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11Sample Output
3 2 2 2 6 1 2 3 4 5 6
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <map> using namespace std; int m,n,s[200000],ans[200000];///ans存要求的序列 int main() { while(scanf("%d",&m) != EOF) { map<int,int> q; n = 0; for(int i = 0;i < m;i ++) { scanf("%d",&s[i]); q[s[i]] ++; } sort(s,s + m); for(int i = 0;i < m;i ++) { if(!q[s[i]])continue;///次数为0，就略过 for(int j = 0;j < n;j ++)///依次跟ans里的值相加来消除后边的数 { q[s[i] + ans[j]] --; } ans[n ++] = s[i]; q[s[i]] --;//防止重复的数读进去，需把次数-1 } printf("%d\n",n); for(int i = 0;i < n;i ++) { if(i)putchar(‘ ‘); printf("%d",ans[i]); } putchar(‘\n‘); } }
标签：cat lines MIXED cli 3.5 nal ram eof and
原文地址：https://www.cnblogs.com/8023spz/p/9002306.html
