HDU 3555(数位dp)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 20122    Accepted Submission(s): 7508

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

3150500

 

Sample Output

0115
题 意 :找出含有49 的数的个数。
思 路: 数位dp的模板题。
代码1： 
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 22using namespace std;long long dp[N][10];//dp[i][j]表示i位数以j开头中不含有49 的个数。int num[N];long long dfs(int pos,int pre,int lim){if(pos<=0) return 1;if(!lim&&dp[pos][pre]!=-1) return dp[pos][pre];int end=lim?num[pos]:9;long long fin=0;for(int i=0;i<=end;i++){if(pre==4&&i==9) continue;fin+=dfs(pos-1,i,lim&&i==end);}if(!lim) dp[pos][pre]=fin;return fin;}void solve(long long x){long long xx=x;int len=0;while(x){num[++len]=x%10;x/=10;}long long fin=dfs(len,0,1);printf("%lld\n",xx-fin+1);//加1   是因为将 0 算入dp其中。return ;}int main(){int t;scanf("%d",&t);memset(dp,-1,sizeof(dp));while(t--){long long n;scanf("%lld",&n);solve(n);}return 0;}

代码2; 

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;long long dp[22][10][2];//dp[i][j][k]表示i位数中以j开头 中 type表示数是否含有 49 。int num[22];long long dfs(int pos,int pre,int type,int lim){if(pos<=0) return type;if(!lim&&dp[pos][pre][type]!=-1) return dp[pos][pre][type];int end=lim?num[pos]:9;long long fin=0;for(int i=0;i<=end;i++){fin+=dfs(pos-1,i,type||pre==4&&i==9,lim&&i==end);}if(!lim) dp[pos][pre][type]=fin;return fin;}void solve(long long x){int len=0;while(x){num[++len]=x%10;x/=10;}long long fin=dfs(len,0,0,1);printf("%lld\n",fin);return ;}int main(){int t;scanf("%d",&t);memset(dp,-1,sizeof(dp));while(t--){long long n;scanf("%lld",&n);solve(n);}return 0;}