XDOJ_1030_二分

Posted 2020-08-15

http://acm.xidian.edu.cn/problem.php?id=1030

 

直接枚举前两个数，二分找第三个。

也可以先把正负分开，分两部分做，速度快了些，但要保存每个答案，耗内存。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int a[1005],n;

int main()
{
    while(~scanf("%d",&n))
    {
        int flag = 1;
        for(int i = 1;i <= n;i++)   scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        for(int i = 1;i <= n-2;i++)
        {
            for(int j = i+1;j <= n-1;j++)
            {
                int sum = -a[i]-a[j];
                int pos = lower_bound(a+j+1,a+n+1,sum)-a;
                if(pos <= n && a[pos] == sum)
                {
                    flag = 0;
                    printf("%d %d %d\n",a[i],a[j],a[pos]);
                }
            }
        }
        if(flag)    printf("Ren Chou Jiu Gai Duo Du Shu!\n");
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

struct anss
{
    int a,b,c;
}ans[1000000];
int a[1005],n;

bool cmp(anss x,anss y)
{
    if(x.a == y.a)  return x.b < y.b;
    return x.a <y.a;
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i = 1;i <= n;i++)   scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        int pos = lower_bound(a+1,a+1+n,0)-a;
        int cnt = 0;
        for(int i = 1;i < pos;i++)
        {
            for(int j = i+1;j < pos;j++)
            {
                int sum = -a[i]-a[j];
                int x = lower_bound(a+pos,a+1+n,sum)-a;
                if(x <= n && a[x] == sum)
                {
                    ans[cnt].a = a[i];
                    ans[cnt].b = a[j];
                    ans[cnt].c = a[x];
                    cnt++;
                }
            }
        }
        for(int i = pos;i <= n;i++)
        {
            for(int j = i+1;j <= n;j++)
            {
                int sum = -a[i]-a[j];
                int x = lower_bound(a+1,a+pos,sum)-a;
                if(x < pos && a[x] == sum)
                {
                    ans[cnt].a = a[x];
                    ans[cnt].b = a[i];
                    ans[cnt].c = a[j];
                    cnt++;
                }
            }
        }
        sort(ans,ans+cnt,cmp);
        if(cnt == 0)    printf("Ren Chou Jiu Gai Duo Du Shu!\n");
        for(int i = 0;i < cnt;i++)  printf("%d %d %d\n",ans[i].a,ans[i].b,ans[i].c);
    }
    return 0;
}