hdu3555 数位dp

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 13181 Accepted Submission(s): 4725

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

3 1 50 500

Sample Output

0 1 15

求1~n中有多少个数中没有49连续子序列的。

思路:

比较简单的数位dp。dp[len][w][is4]表示第len位的时候，之前是否存在49,并且之前的数字是否为4.

dp[len][w][is4] = sum(dp[len-1][fw][fis4]);

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 50;
ll dp[MAXN][2][2];
int digit[MAXN];
char s[MAXN];
ll dfs(int len,int w,int ismax,int pa,int is4)
{
    if(len == 0)return w ? 1 : 0;
    if(!ismax && dp[len][w][is4])return dp[len][w][is4];
    int maxv = ismax ? digit[len] : 9;
    ll ans = 0;
    for(int i = 0; i <= maxv; i++){
        if(pa == 4 && i == 9){
            ans += dfs(len-1,1,ismax && i == maxv,i,i == 4);
        }
        else {
            ans += dfs(len-1,w,ismax && i == maxv,i,i == 4);
        }
    }
    if(!ismax)dp[len][w][is4] = ans;
    return ans;
}
void solve()
{
    int slen = strlen(s);
    int len = 0;
    for(int i = slen - 1; i >= 0; i--){
        digit[++len] = s[i] - '0';
    }
    memset(dp,0,sizeof(dp));
    printf("%lld
",dfs(len,0,1,-1,0));
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",s);
        solve();
    }
    return 0;
}