hdu-6168 Numbers

Numbers

题意：给一组数，包括了a数组和b数组，并且已知b数组是由a数组里的数两两相加得到的，求a数组。

思路：想到输入的数组里的最小的两个数肯定不在b数组里，即在a数组里，所以：我们先把最小的两个数放进a数组，然后将两个数组相加得到一个数，把这个数去将输入的数组里的一样的数去掉一个，然后再从里面调一个数出来，将它与a数组里所有的元素相加，得到一些数，然后用这些书去掉输入数组里的数。最后就可以得到a数组了。

#include <cstdio>#include <iostream>#include <string.h>#include <string>#include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <stack>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))#define LL long long#define siz 2505using namespace std;int m;int gp[125255];//int p[125255];int ans[125255];//int tx[125255];void solve(){    priority_queue<int,vector<int>,greater<int> > que;    int n = 0;    int s = 1,e = 2;    int zx = 0,i = 3,op = 1;    //p[++zx] = gp[s] + gp[e];    que.push(gp[s] + gp[e]);    ans[++n] = gp[s];    ans[++n] = gp[e];   // cout<<1<<endl;  // int tt = 0;   //tx[++tt] = gp[s],tx[++tt] = gp[e];   s = e;    while(i<=m){       // cout<<i<<"----"<<op<<endl;        while(i<=m&&!que.empty()){            int u = que.top();           // cout<<op<<" "<<i<<" "<<gp[i]<<" "<<u<<" "<<endl;            if(u == gp[i]){            ++i; que.pop();}            else break;        }        if(i>m) break;        //cout<<op<<endl;       // s = i;        //ans[++n] = gp[i];        //cout<<"---------"<<gp[i]<<"-----------"<<endl;        for(int j = 1;j<=n;j++){            //p[++zx] = ans[j] + gp[i];            que.push(ans[j] + gp[i]);            //cout<<ans[j] + gp[i]<<" -----------"<<endl;        }        //cout<<n<<"&&&&&&&&&&&&&&&&&&&&&"<<i<<" "<<gp[i]<<endl;        ans[++n] = gp[i];        ++i;    }    printf("%d\n%d",n,ans[1]);    for(int i=2;i<=n;i++){        printf(" %d",ans[i]);    }    printf("\n");}int main(){    while(~scanf("%d",&m)){        for(int i=1;i<=m;i++){            scanf("%d",&gp[i]);        }        sort(gp+1,gp+m+1);        solve();    }    return 0;}