标签：RKE 一个 lag imp lib ace NPU case \n
Bomb HDU - 3555 （数位DP）
InputThe 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.
OutputFor 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.
题意：求1-n之间有几个数带有49
题解：一题数位DP，可以用记忆化搜索解决，感觉板子还是有点懂了，但不知道改了之后会不会了，具体的dfs的每一步的具体操作已经放在代码里了
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;
using namespace std;
#define ms(a,b) memset(a,b,sizeof(a))
#define lson rt*2,l,(l+r)/2
#define rson rt*2+1,(l+r)/2+1,r
typedef unsigned long long ull;
typedef long long ll;
const int MAXN=1e4+5;
const double EPS=1e-8;
const int INF=0x3f3f3f3f;
const int MOD = 1e9+7;
#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = 1e5+5;
const int mod = 1e9+7;
int bit[40];
ll f[40][4];
ll dp(int pos,int st,bool flag)
{
//pos为位，st为状态，st=0：没有49，st=1：前一位为4，st=2，表示49都已经出现过了
//flag表示高位与原数是否相同
if(pos == 0)
return st == 2;
if(flag && f[pos][st] != -1)
return f[pos][st];
ll ans = 0;
int x;
if(flag)
x = 9;
else
x = bit[pos];
//cout<<x<<endl;
for(int i = 0;i <= x; i++)
{
if((st == 2) || (st == 1 && i == 9)) //如果上一位已经有49（st==2）或者上一位为4（st==1）当前位为9
ans += dp(pos-1,2,flag || i<x); //只能加上上一个状态，且上一个状态一定为已经有49了
else if(i == 4)
ans += dp(pos-1,1,flag || i<x); //当前为4和前一位也为4的状态是一样的
else //维持st=0的状态
ans += dp(pos-1,0,flag || i<x);
}
if(flag)
f[pos][st] = ans; //记忆化
return ans;
}
ll calc(ll x)
{
int len = 0;
while(x)
{
bit[++len] = x % 10;
x /= 10;
}
//cout<<len<<endl;
return dp(len,0,0);
}
int main()
{
int t;
scanf("%d",&t);
memset(f,-1,sizeof f);
while(t--)
{
ll n;
scanf("%lld",&n);
printf("%lld\n",calc(n));
}
}
标签：RKE 一个 lag imp lib ace NPU case \n
原文地址：https://www.cnblogs.com/smallhester/p/10324872.html
