HDU 3555 数位DP

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define LL long long
//#define __int64 LL
LL dp[22][3];
int pos[22];
//[][0] 包含49
//[][1] 最高位是9 但不含49
//[][2] 不含49
void initial()
{
    dp[0][2]=1;
    for(int i = 1; i < 22; i++)
    {
        dp[i][0] = dp[i-1][0] * 10 + dp[i-1][1];
        dp[i][1] = dp[i-1][2];
        dp[i][2] = dp[i-1][2] * 10 - dp[i-1][1];
    }
}
LL ans(LL a)
{
    int i,cnt = 0;
    LL res = 0;
    memset(pos,0,sizeof(pos));
    a++;
    while(a>10)
    {
        pos[cnt++] = a % 10;
        a /= 10;
    }
    pos[cnt] = a;
    bool flag = false;
    for(i = cnt; i >= 0; i--)
    {
        //cout<<dp[i][0]<<" "<<dp[i][1]<<" "<<dp[i][2]<<endl;
        res += pos[i]*dp[i][0];
        if(flag)
            res += dp[i][2] * pos[i];
        if(!flag && pos[i] > 4 )
            res += dp[i][1];
        if(pos[i+1] == 4 && pos[i] == 9)
            flag = true;
    }
    return res;
}
int main()
{
    int n;
    cin>>n;
    initial();
    while(n--)
    {
        LL a;
        cin>>a;
        cout<<ans(a)<<endl;
    }
    return 0;
}

没什么好说的水题一枚...虽然我也不是很会做也是看题解才懂得

第一题数位DP...mark一下咯

来源：https://www.cnblogs.com/Felix-F/archive/2013/04/02/3223633.html