ACM 第一天

标签库元素：

队列<queue> FIFO

栈 <stack>  FICO

集合 set

不定长数组  vector

映射 map

 

Maximum Multiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3985    Accepted Submission(s): 926

Problem Description

 

Given an integer n, Chiaki would like to find three positive integers x, y and z such that: n=x+y+z, x∣n, y∣n, z∣n and xyz is maximum.

 

 

Input

 

There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤106).

 

 

Output

 

For each test case, output an integer denoting the maximum xyz. If there no such integers, output −1 instead.

 

 

Sample Input

 

3 1 2 3

 

 

Sample Output

 

-1 -1 1

#include<iostream>
#include<vector>
#include<stdio.h>
using namespace std;

int main()
{
    long long n;
    int t ;
    cin>>t;
    while(t--)
    {
        scanf("%lld",&n);
        if(n%3 == 0) cout<<(n/3)*(n/3)*(n/3)<<endl;
        else
        {
            if(n%4 == 0) cout<<(n/2)*(n/4)*(n/4)<<endl;
            else cout<<-1<<endl;
        }

    }
    return 0;
}

 

Triangle Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 2140    Accepted Submission(s): 925
Special Judge

Problem Description

Chiaki has 3n points p1,p2,…,p3n. It is guaranteed that no three points are collinear.
Chiaki would like to construct n disjoint triangles where each vertex comes from the 3n points.

 

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤1000) -- the number of triangle to construct.
Each of the next 3n lines contains two integers xi and yi (−109≤xi,yi≤109).
It is guaranteed that the sum of all n does not exceed 10000.

 

 

Output

For each test case, output n lines contain three integers ai,bi,ci (1≤ai,bi,ci≤3n) each denoting the indices of points the i-th triangle use. If there are multiple solutions, you can output any of them.

 

 

Sample Input

1 1 1 2 2 3 3 5

 

 

Sample Output

1 2 3

#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
struct P
{
    long long x,y;
    int id;
}p[3000+10];
int cmp(P a,P b)
{
    return a.x<b.x;
}

int main()
{

    int t;
    cin>>t;
    while(t--)
    {
        int n ;
        cin>>n;
        for(int i = 1;i <= 3*n;i++)
        {
            scanf("%lld %lld",&p[i].x,&p[i].y);
            p[i].id = i;
        }
        sort(p+1,p+3*n+1,cmp);

        for(int i = 1;i<=3*n;i++)
        {

            cout<<p[i].id;
            if(i%3 == 0) cout<<endl;
            else cout<<" ";
        }
    }
    return 0;
}

 

 

Time Zone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5204    Accepted Submission(s): 878

Problem Description

Chiaki often participates in international competitive programming contests. The time zone becomes a big problem.
Given a time in Beijing time (UTC +8), Chiaki would like to know the time in another time zone s.

 

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:
The first line contains two integers a, b (0≤a≤23,0≤b≤59) and a string s in the format of "UTC+X'', "UTC-X'', "UTC+X.Y'', or "UTC-X.Y'' (0≤X,X.Y≤14,0≤Y≤9).

 

 

Output

For each test, output the time in the format of hh:mm (24-hour clock).

 

 

Sample Input

3 11 11 UTC+8 11 12 UTC+9 11 23 UTC+0

 

 

Sample Output

11:11 12:12 03:23

#include <bits/stdc++.h>
#include <vector>
#include <queue>

using namespace std;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int a,b;
        char f;
        double k;
        int q=0;
        scanf("%d %d UTC%c%lf",&a,&b,&f,&k);
        k=k*10;
       int m1=((int)k%10)*6;
       int  h1=(int)k/10;
       a=(a-8+24)%24;

        if(f=='+')
        {
            if(b+m1>=60)
           {
            q=1;
           }
           else q=0;
           m1=(b+m1)%60;
           h1=(a+q+h1)%24;
        }
       else if(f=='-')
         {
            if(m1>b)
           {
              q=1;
           }
            else q=0;

           m1=(b+60-m1)%60;

           h1=(a-q-h1+24)%24;

         }
         if(h1<=9)
         {
              printf("0%d:",h1);
         }
         else printf("%d:",h1);

         if(m1<=9)
         {
              printf("0%d\n",m1);
         }
         else printf("%d\n",m1);

        }


    return 0;
}