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HDU 6020---MG loves apple（枚举）

2024-09-12 03:15:49   215人阅读

题目链接

 

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
 
题意：

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思路：

技术分享

 

代码如下：

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;int  a[100005];char s[100005];void cal(int &a3, int &E1,int &E2,int N){    a3=0; E1=0; E2=0;    for(int i=1;i<=N;i++)    {        if(a[i]==3) break;        if(a[i]==0) a3++;    }    for(int i=1;i<=N;i++)    {        if(a[i]==0) break;        if(a[i]==1) E1=1;        if(a[i]==2) E2=1;    }    return ;}int main(){    int T;    cin>>T;    while(T--)    {        int N,K;        int s1=0,s2=0,s3=0;        scanf("%d%d",&N,&K);        scanf("%s",s+1);        for(int i=1;i<=N;i++)        {            a[i]=s[i]-‘0‘;            if(a[i]%3==1) a[i]=1,s1++;            else if(a[i]%3==2) a[i]=2,s2++;            else s3++,a[i]=(a[i])?3:0;        }        int ans=(s1+s2*2)%3;        int a3,E1,E2,f=0;        cal(a3,E1,E2,N);        for(int C=0;C<=s2&&C<=K;C++)  ///C->2; B->1; A->0;        {            int B=((ans-C*2)%3+3)%3;            for(;B<=s1&&C+B<=K;B=B+3)            {                int A=K-C-B;                if(A<=s3)                {                    if(A>a3) f=1;                    else if(B<s1&&E1) f=1;                    else if(C<s2&&E2) f=1;                    if(f) break;                }            }            if(f) break;        }        if((N==K+1)&&s3) f=1;        if(f) puts("yes");        else puts("no");    }    return 0;}

 

 

HDU 6020---MG loves apple（枚举）


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