hdu 6020 MG loves apple 恶心模拟

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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

BestCoder Round #93

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 ;
}