HDU6020 MG loves apple

最新推荐文章于 2018-08-28 13:54:56 发布

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探讨了如何从一系列苹果中移除特定数量的苹果，使得剩余苹果组成的数字不含前导零且能被3整除的问题。通过统计苹果价值模3的结果来寻找可行解。

MG loves apple

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 442 Accepted Submission(s): 76

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

Sample Output

  
   yes
no

Source

BestCoder Round #93

官方题解：

这题更多算是想法题，有脑洞的同学可以各种乱搞。将问题转化成：求去掉 $K$ 位数字后，不含前导零，且数字和是否能被三整除。按照预测，这里应该会汇聚本场最多的hack点。

我们设 $S 0$ 、 $S 1$ 、 $S 2$ 分别为原串上 $m o d 3 = 0$ 、 $1$ 、 $2$ 数字的个数。我们假定删除取模后为 $0$ 、 $1$ 、 $2$ 的数字各 $A$ 、 $B$ 、 $C$ 个，则显然有 $0 < = A < = S 0, 0 < = B < = S 1, 0 < = C < = S 2$ 且 $K = A + B + C$ 且 $S u m$ $m o d 3 = (A * 0 + B * 1 + C * 2) m o d 3 = (S 0 * 0 + S 1 * 1 + S 2 * 2) m o d 3 = b i a s$ 。枚举 $C$ 的值，我们可得 $B m o d 3 = (b i a s - C * 2) m o d 3, A = K - B - C$ 。如果有若干组 $A, B$ 不逾界，可知这些 $(A, B, C)$ 是在模意义下合法的解，但不一定满足没有前导零。

所以，对于【大于 $0$ 的数】我们贪心地从后往前删除，对于 $0$ 我们贪心地从前往后删除。

需要统计出： $a 3$ =第一个【 $m o d 3 = 0$ 且非 $0$ 的数】前 $0$ 的个数（如果 $m o d 3 = 0$ 且非 $0$ 的数不存在，那么 $a 3$ 就取所有零的个数）， $E 1$ =【第一个 $0$ 前是否存在 $m o d 3 = 1$ 的数】， $E 2$ =【第一个 $0$ 前是否存在 $m o d 3 = 2$ 的数】。

则以下情况满足任一种都能保证无前导零： $A > = a 3$ 。 $B <$ $S 1$ 且 $E 1$ 。 $C <$ $S 2$ 且 $E 2$ 。

还需要加一种情况满足k==n-1 也是yes

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<string.h>
#include<math.h>
#include<list>

using namespace std;

#define ll long long
#define pii pair<int,int>
const int inf = 1e9 + 7;

const int N = 100000 + 5;
char ch[N];
int x[N];

bool slove(int n,int k){
    for(int i=0;i<n;++i){
        x[i]=ch[i]-'0';
        x[i]%=3;
    }
    int s[3]={0,0,0};
    int sum=0;
    for(int i=0;i<n;++i){
        s[x[i]]+=1;
        sum+=x[i];
    }
    int a3=0;//>0&&mod3==0的第一数 前面的0的个数
    bool e1,e2=e1=0;
    for(int i=0;i<n;++i){
        if(ch[i]!='0'&&x[i]==0){
            break;
        }
        a3+=(x[i]==0);
        if(x[i]==1){
            e1=1;
        }
        if(x[i]==2){
            e2=1;
        }
    }
    for(int c=0;c<=s[2];++c){
        int minb=((sum-2*c)%3 + 3)%3;
        for(int b=minb;b<=s[1];b+=3){
            int a=k-b-c;
            if(a>=0&&a<=s[0]){
                if((a>=a3)||(e1&&b<s[1])||(e2&&c<s[2])){
                    return 1;
                }
                if(k==n-1){
                    return 1;
                }
            }

        }
    }
    return 0;
}

int main()
{
    //freopen("/home/lu/Documents/r.txt","r",stdin);
    //freopen("/home/lu/Documents/w.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--){
        int n,k;
        scanf("%d%d%s",&n,&k,ch);
        printf("%s\n",slove(n,k)?"yes":"no");
    }
    return 0;
}