hdu 6020 MG loves apple 恶心模拟

时间：2021-02-26 01:49:06

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MG loves apple

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Problem Description
MG is a rich boy. He has n apples, each has a value of V(0<=V<=9).

A valid number does not contain a leading zero, and these apples have just made a valid N digit number.

MG has the right to take away K apples in the sequence, he wonders if there exists a solution: After exactly taking away K apples, the valid N−K digit number of remaining apples mod 3 is zero.

MG thought it very easy and he had himself disdained to take the job. As a bystander, could you please help settle the problem and calculate the answer?

 
Input
The first line is an integer T which indicates the case number.（1<=T<=60)

And as for each case, there are 2 integer N(1<=N<=100000),K(0<=K<N) in the first line which indicate apple-number, and the number of apple you should take away.

MG also promises the sum of N will not exceed 1000000。

Then there are N integers X in the next line, the i-th integer means the i-th gold’s value(0<=X<=9).

 
Output
As for each case, you need to output a single line.

If the solution exists, print”yes”,else print “no”.(Excluding quotation marks)

 
Sample Input
2
5 2
11230
4 2
1000
 
Sample Output
yes
no
 
Source
思路：枚举以i+1为起点的，去掉前i个，后面的需要去掉k-i个，进行check（）
#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<iostream>

#include<cstdio>

#include<cmath>

#include<string>

#include<queue>

#include<algorithm>

#include<stack>

#include<cstring>

#include<vector>

#include<list>

#include<set>

#include<map>

using namespace std;

#define ll long long

#define pi (4*atan(1.0))

#define eps 1e-14

#define bug(x)  cout<<"bug"<<x<<endl;

const int N=1e5+,M=1e6+,inf=;

const ll INF=1e18+,mod=;

int num[N][];

int check0(int a,int x,int y,int z)

{

    if(a<||x<||y<||z<)return ;

    int ans=;

    if(x>=&&y>=&&z>=)

    {

        int xx=x,yy=y,zz=z;

        xx-=;yy-=;zz-=;

        int l=(xx/+yy/);

        zz-=min(zz/,l)*;

        if(zz<=a)ans=;

    }

    if(x>=&&y>=&&z>=)

    {

        int xx=x,yy=y,zz=z;

        xx-=;yy-=;zz-=;

        int l=(xx/+yy/);

        zz-=min(zz/,l)*;

        if(zz<=a)ans=;

    }

    int xx=x,yy=y,zz=z;

    int l=(xx/+yy/);

    zz-=min(zz/,l)*;

    if(zz<=a)ans=;

    return ans;

}

int check1(int a,int x,int y,int z)

{

    return max(check0(a,x-,y,z-),check0(a,x,y-,z-));

}

int check2(int a,int x,int y,int z)

{

    return max(check0(a,x,y-,z-),check0(a,x-,y,z-));

}

char a[N];

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        memset(num,,sizeof(num));

        int n,K;

        scanf("%d%d",&n,&K);

        scanf("%s",a+);

        if(K==n-)

        {

            int mmp=;

            for(int i=;i<=n;i++)

            if(a[i]%==)mmp=;

            if(mmp)printf("yes\n");

            else printf("no\n");

            continue;

        }

        int sum=;

        for(int i=;i<=n;i++)

            sum+=a[i]-'';

        int x=sum%;

        for(int i=n;i>=;i--)

        {

            int x=(a[i]-'')%;

            num[i][]=num[i+][];

            num[i][]=num[i+][];

            num[i][]=num[i+][];

            num[i][x]++;

        }

        int ans=;

        /// 枚举以i+1为起点的，去掉前i个，后面的需要去掉k-i个，进行check（）

        for(int i=;i<=K;i++)

        {

            if(a[i+]=='')continue;

            if(x==)ans=max(ans,check1(num[i+][],num[i+][],num[i+][],K-i));

            else if(x==)ans=max(ans,check2(num[i+][],num[i+][],num[i+][],K-i));

            else ans=max(ans,check0(num[i+][],num[i+][],num[i+][],K-i));

            x-=a[i+]-'';

            x=(x%+)%;

        }

        if(ans)

            printf("yes\n");

        else

            printf("no\n");

    }

    return ;

}
 

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