Bomb(HDU 3555 数位DP)

最新推荐文章于 2021-08-09 21:12:02 发布

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#Bomb #HDU 3555 #数位DP

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本文介绍了一道关于寻找1到N内包含特定数字序列49的问题，并提供了一个使用数位DP方法的解决方案。该方法通过构建状态转移方程来高效地计算答案。

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Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 20230 Accepted Submission(s): 7565

Problem Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

Sample Output

0
1
15

HintFrom 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",
so the answer is 15.

//题意：给一个数字n，问1-n里有几个数包含49.

//思路：数位DP：

dp[i][0]代表长度为 i 并且不含有49的数字的个数；

dp[i][1]代表长度为 i 并且不含有49，但是最高位是9的数字的个数；

dp[i][2]代表长度为 i 并且含有49的数字的个数。

状态转移方程：

dp[i][0] = dp[i-1][0] * bit[i] - dp[i-1][1];

dp[i][1]=dp[i-1][1];

dp[i][2]=dp[i-1][2]*bit[i]+dp[i-1][1];

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

ll n;
int bit[25];
ll dp[25][5];

 int main()
{
    int T;
    scanf("%d",&T);
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(int i=1;i<21;i++)
    {
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
        dp[i][1]=dp[i-1][0];
        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
    }
    while(T--)
    {
        scanf("%I64d",&n);
        memset(bit,0,sizeof(bit));
        ll x=n+1;
        int len=0;
        while(x)
        {
            bit[++len]=x%10;
            x=x/10;
        }
        ll ans=0;
        bool flag=false;
        int last=0;
        for(int i=len;i>=1;i--)
        {
            ans+=(dp[i-1][2]*bit[i]);
            if(flag)
                ans+=dp[i-1][0]*bit[i];
            if(!flag&&bit[i]>4)
                ans+=dp[i-1][1];
            if(last==4&&bit[i]==9)
                flag=true;
            last=bit[i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}