【HDU3555】数位Dp1~n之间出现特征数字个数

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 7279    Accepted Submission(s): 2541


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

3
1
50
500

 

Sample Output

0
1
15

HintFrom 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

2010
ACM-ICPC Multi-University Training Contest（12）——Host by WHU

转自：http://blog.csdn.net/winkloud/article/details/7866520

做的第一道数位DP啊！开始在找规律，搜索，做了很久终于找到了规律，上网一查发现原来这样的叫数位DP。。


找到的规律就是这个样子了。有了规律就很好做了。dp[i][0]=dp[i-1][0]*10-dp[i-1][1];是因为要减去49XXX的情况。

题意就是找0到n有多少个数中含有49。数据范围接近10^20

DP的状态是2维的dp[len][3]

dp[len][0] 代表长度为len不含49的方案数

dp[len][1] 代表长度为len不含49但是以9开头的数字的方案数

dp[len][2] 代表长度为len含有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]

//Time:15MS
//Memory:488K
#include<string.h>
#include<stdio.h>
long long dp[20][3];
int num[20];
int main()
{
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] 表示i位数字中不含49的数字的个数
dp[i][1]=dp[i-1][0];               //dp[i][1] 表示i位数字中以9开头的数字的个数
dp[i][2]=dp[i-1][2]*10+dp[i-1][1];//dp[i][2] 表示i位数字中含有49的数字的个数
}
int t;
scanf("%d",&t);
while(t--)
{
int len = 0,last = 0;
long long ans = 0;
long long n = 0;
scanf("%I64d",&n);
n++;
memset(num,0,sizeof(num));
while(n){
num[++len]=n%10;
n/=10;
}
bool flag=false;
for(int i=len;i>=1;i--)
{
ans+=dp[i-1][2]*num[i];
if(flag)
{
ans+=dp[i-1][0]*num[i];
}
if(!flag && num[i]>4)
{
ans+=dp[i-1][1];
}
if(last==4 && num[i]==9)
{
flag=true;
}
last=num[i];
}
printf("%I64d\n",ans);
}
}