BestCoder Round #85 前三题题解

sum

Accepts: 822

Submissions: 1744

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO

Input

The first line of the input has an integer T (1≤T≤101 \leq T \leq 101≤T≤10), which represents the number of test cases. For each test case, there are two lines: 1.The first line contains two positive integers n, m (1≤n≤1000001 \leq n \leq 1000001≤n≤100000, 1≤m≤50001 \leq m \leq 50001≤m≤5000). 2.The second line contains n positive integers x (1≤x≤1001 \leq x \leq 1001≤x≤100) according to the sequence.

Output

Output T lines, each line print a YES or NO.

Sample Input

2
3 3
1 2 3
5 7
6 6 6 6 6

Sample Output

YES
NO

Statistic | Submit | Clarifications | Back

 

第一题:同余,因为要求的是 (sum[i]-sum[j])%m == 0 ，所以 sum[i]%m = sum[j]%m ,所以判断一下所有的前缀和里面是否有余数相同的两个就好了,还要判一下sum[i]%m==0也是可行的.

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
#include <queue>
#include <stdlib.h>
using namespace std;
typedef long long LL;
const int N = ;
int sum[N];
bool vis[];
int main()
{
    int tcase,n,m,x;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d%d",&n,&m);
        sum[] = ;
        bool flag = false;
        memset(vis,false,sizeof(vis));
        for(int i = ;i<=n;i++){
            scanf("%d",&x);
            sum[i] = sum[i-]+x;
            if(sum[i]%m==) flag = true;
            if(vis[sum[i]%m]) flag = true;
            vis[sum[i]%m] = true;
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return ;
}

domino

Accepts: 614

Submissions: 1498

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Little White plays a game.There are n pieces of dominoes on the table in a row. He can choose a domino which hasn't fall down for at most k times, let it fall to the left or right. When a domino is toppled, it will knock down the erect domino. On the assumption that all of the tiles are fallen in the end, he can set the height of all dominoes, but he wants to minimize the sum of all dominoes height. The height of every domino is an integer and at least 1.

Input

The first line of input is an integer T ( 1≤T≤101 \leq T \leq 10 1≤T≤10) There are two lines of each test case. The first line has two integer n and k, respectively domino number and the number of opportunities.( 2≤k,n≤1000002\leq k, n \leq 100000 2≤k,n≤100000) The second line has n - 1 integers, the distance of adjacent domino d, 1≤d≤1000001 \leq d \leq 100000 1≤d≤100000

Output

For each testcase, output of a line, the smallest sum of all dominoes height

Sample Input

1
4 2
2 3 4

Sample Output

9

Statistic | Submit | Clarifications | Back

 

题解:贪心,取前K大的距离减掉，其余的化为距离+1，最后一块为1，单独处理 k>=n的情况

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
#include <queue>
using namespace std;
const int N = ;
int a[N];
int cmp(int a,int b){
    return a>b;
}
int main(){
    int tcase,n,k;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d%d",&n,&k);
        long long sum = ;
        for(int i=;i<n;i++){
            scanf("%d",&a[i]);
            sum=sum+a[i]+;
        }
        if(n<=k){
            printf("%d\n",n);
            continue;
        }
        sort(a+,a+n,cmp);
        for(int i=;i<k;i++){
            sum-=a[i];
        }
        printf("%I64d\n",sum+);
    }
    return ;
}

abs

Accepts: 241

Submissions: 927

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Given a number x, ask positive integer y≥2y\geq 2y≥2, that satisfy the following conditions:

1. The absolute value of y - x is minimal
2. To prime factors decomposition of Y, every element factor appears two times exactly.

Input

The first line of input is an integer T ( 1≤T≤501\leq T \leq50 1≤T≤50) For each test case，the single line contains, an integer x ( 1≤x≤10181\leq x \leq {10} ^ {18} 1≤x≤10​18​​)

Output

For each testcase print the absolute value of y - x

Sample Input

5
1112
4290
8716
9957
9095

Sample Output

23
65
67
244
70

题解:分别往左找往右找,如果这个数分解的质因数个数都只有1，那么他本身的平方每个质因数个数就为2了，往右的时候(k+i)*(k+i)>=n这个条件一定要记得写。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
#include <queue>
#include <stdlib.h>
using namespace std;
typedef long long LL;
const LL INF = (LL)(((LL))<<-);
bool can(LL num)
{
    for(int i=; i*i<=num; i++)
    {
        int cnt = ;
        if(num%i==)
        {
            while(num%i==)
            {
                num/=i;
                cnt++;
                if(cnt>) return false;
            }
        }
    }
    return true;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {

        LL n;
        scanf("%I64d",&n);
        if(n<=){
            printf("%I64d\n",-n);
            continue;
        }
        LL k = sqrt(n);
        LL ans = INF;
        for(int i=;; i++)
        {
            if((k+i)*(k+i)>=n&&can(k+i))
            {
                ans = min(ans,abs((k+i)*(k+i)-n));
                break;
            }
        }
        for(int i=;; i++)
        {
            if(can(k-i)&&(k-i)>=)
            {
                ans = min(ans,abs((k-i)*(k-i)-n));
                break;
            }
        }
        printf("%I64d\n",ans);
    }
    return ;
}