hdu3555 Bomb[数位dp]

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 17734 Accepted Submission(s): 6522

　

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
Hint

From 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.

Author

fatboy_cw@WHU

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

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数位dp

数位dp的难度不在于转移，要算出一个整百整千整万的数包含多少个是很容易的（转移很简单），但是我们给的数都是形如“遡遢遣遤遥遦”这种数……零头的计算往往包含很多细节，而记忆化搜索很好的解决这个问题！！

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const int N=20;
ll n,f[N][3];int T,bits[N];
ll dfs(int pos,int st,bool lim){
    if(!pos) return st==2;
    ll &res=f[pos][st],ans=0;
    if(!lim&&(~res)) return res;
    int up=!lim?9:bits[pos];
    for(int i=up;~i;i--){
        if(st==2||(st==1&&i==9))
            ans+=dfs(pos-1,2,lim&&i==bits[pos]);
        else if(i==4)
            ans+=dfs(pos-1,1,lim&&i==bits[pos]);
        else 
            ans+=dfs(pos-1,0,lim&&i==bits[pos]);
    }
    if(!lim) res=ans;
    return ans;
}
ll solve(ll x){
    int len=0;
    for(;x;x/=10) bits[++len]=x%10;
    return dfs(len,0,1);
}
int main(){
    memset(f,-1,sizeof f);
    for(scanf("%d",&T);T--;){
        scanf("%lld",&n);
        printf("%lld
",solve(n));
    }
    return 0;
}