hdu3555 Bomb(数位DP)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7209    Accepted Submission(s): 2518

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
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:给一个N，问1-N包含了多少个49。N为2的63次方-1.

数位DP思想

• //dp[i][0]代表长度为i不含49的方案数 
• //dp[i][1]代表长度为i高位为9的方案数 
• //dp[i][2]代表长度为i含49的方案数 
• 状态转移如下
• dp[i][0] = dp[i-1][0] * 10 - dp[i-1][1];  // not include 49  如果不含49且，在前面可以填上0-9 但是要减去dp[i-1][1] 因为4会和9构成49
• dp[i][1] = dp[i-1][0];  // not include 49 but starts with 9  这个直接在不含49的数上填个9就行了
• dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1]; // include 49  已经含有49的数可以填0-9，或者9开头的填4
• 接着就是从高位开始统计
• 在统计到某一位的时候，加上 dp[i-1][2] * digit[i] 是显然对的，因为这一位可以填 0 - (digit[i]-1)
• 若这一位之前挨着49，那么加上 dp[i-1][0] * digit[i] 也是显然对的。
• 若这一位之前没有挨着49，但是digit[i]比4大，那么当这一位填4的时候，就得加上dp[i-1][1]

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <queue>#include <map>#include <stack>#include <list>#include <vector>using namespace std;#define LL __int64LL dp[30][5];LL f[30];void INIT(){memset(dp,0,sizeof(dp));dp[0][0]=1;for (int i=1;i<=20;i++){dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//dp[i][0]代表长度为i不含49的方案数 dp[i][1]=dp[i-1][0];//dp[i][1]代表长度为i高位为9的方案数 dp[i][2]=dp[i-1][1]+dp[i-1][2]*10;//dp[i][2]代表长度为i含49的方案数 }}int main(){int T,i;LL n;INIT();scanf("%d",&T);while(T--){scanf("%I64d",&n);n++;memset(f,0,sizeof(f));int t=0;while (n){f[++t]=n%10;n/=10;}f[t+1]=0;LL ans=0;int flag=0;for (i=t;i>=1;i--){ans+=f[i]*dp[i-1][2];if (flag)ans+=f[i]*dp[i-1][0];elseif (f[i]>4)ans+=dp[i-1][1];if (f[i+1]==4 && f[i]==9)flag=1;}printf("%I64d\n",ans);}return 0;}