hdu3555 Bomb(数位dp)
题目传送门
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 23853 Accepted Submission(s): 8990
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?
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.
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
Recommend
题意:给你n,从[1,n]中找出含有“49”的数的个数
题解:数位dp入门题
代码:
第一份是间接法做的
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
ll n;
int bit[20];
ll dp[20][2];
ll dfs(int pos,int pre,int sta,bool limit)
{
if(pos==-1)return 1;
if(!limit&&dp[pos][sta]!=-1)return dp[pos][sta];
int up=limit?bit[pos]:9;
ll ans=0;
for(int i=0;i<=up;i++)
{
if(pre==4&&i==9)
continue;
ans+=dfs(pos-1,i,i==4,limit&&i==bit[pos]);
}
if(!limit)dp[pos][sta]=ans;
return ans;
}
ll solve(ll x)
{
int len=0;
while(x)
{
bit[len++]=x%10;
x/=10;
}
return dfs(len-1,-1,0,true);
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%lld",&n);
memset(dp,-1,sizeof(dp));
printf("%lld\n",n+1-solve(n));
}
return 0;
}
下面的是直接法做的:
pre=0: 没有49; pre=1: 前一位为4; pre=2: 前几位中有49
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<string.h> #include<algorithm> #include<stdio.h> #include<queue> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef long long ll; typedef pair<int,int> PII; #define mod 1000000007 #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second //head ll n; int bit[20]; ll dp[20][3]; ll dfs(int pos,int pre,bool limit) { if(pos==-1)return pre==2; if(!limit&&dp[pos][pre]!=-1)return dp[pos][pre]; int up=limit?bit[pos]:9; ll ans=0; for(int i=0;i<=up;i++) { if(pre==2||pre==1&&i==9) ans+=dfs(pos-1,2,limit&&i==bit[pos]); else if(i==4) ans+=dfs(pos-1,1,limit&&i==bit[pos]); else ans+=dfs(pos-1,0,limit&&i==bit[pos]); } if(!limit)dp[pos][pre]=ans; return ans; } ll solve(ll x) { int len=0; while(x) { bit[len++]=x%10; x/=10; } return dfs(len-1,0,true); } int main() { int T; scanf("%d",&T); memset(dp,-1,sizeof(dp)); while(T--){ scanf("%lld",&n); printf("%lld\n",solve(n)); } return 0; }