hdu-3555-Bomb

Bomb

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

Total Submission(s): 7781 Accepted Submission(s): 2726



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

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题目大意：给你一个数n叫你求出求出1到n中含有49的数的个数，如500里面有"49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",这16个

思路：
这是做的第一道数位dp的题目，我对数位dp的理解就是对数的每一位只取到该位的整数位（例如4444的话当对第四位进行数位dp时只取到0到4000的数，接着对第三位进行数位dp，这时数位dp的范围是4000到4400的数，所以保证不会有重复，往后依次类推）
数位dp，先预处理一遍，dp[i][0]代表该位为9的所有不合法的状态,dp[i][1]代表该位不合法的所有状态,dp[i][2]代表该位合法的所有状态.然后先预处理
预处理9，99,999...，9999999999..的所有dp[i][j]的情况.dp[i][0] 必须是该位为9的所有不合法的状态，不然的话会有重复的情况如（99999
对第5位进行dp时，这时ans+=dp[i-1][2] + dp[i-1][0],其中dp[i-1][0]的意思是当前位为0下一位为9的情况，如果dp[i][0]代表该位为9的所有状态，那么49490肯定会重复计算，因为49490也被包括在了dp[i-1][2]里面了）

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <queue>
typedef __int64 LL;
using namespace std;
LL dp[100][3];
//dp[i][0]代表该位为9的所有不合法的状态
//dp[i][1]代表该位不合法的所有状态
//dp[i][2]代表该位合法的所有状态
LL n;
void f()
{
int i;
dp[1][0] = 1;
dp[1][1] = 10;
dp[1][2] = 0;
dp[0][1] = 1;
for(i = 2; i <=25; i++)
{
dp[i][0] = dp[i-1][1] ;
dp[i][2] = dp[i-1][0]+ 10*dp[i - 1][2];
dp[i][1] = 10*(dp[i-1][1]) - dp[i-1][0];

}
}
LL ff()
{
LL a[30];
LL k = n;
int len = 1;
while (k > 0)
{
a[len++] = k % 10;
k /= 10;
}
//cout<<len<<endl;
int i;
int vis = 0;
LL ans = 0;
for(i = len - 1; i >= 1; i--)
{
LL m = a[i];
ans += m * dp[i-1][2];
if(vis == 1)
{
ans+= m * (dp[i-1][1]);
//因为数位dp的过程中每次都是以上一个数的最大数作为前缀，所以当该数中某两位
//相邻的出现了49那么接下来的数不管是什么肯定都是合法的。

}
else if(m > 4)
{
ans += dp[i-1][0];
}
if(i!=len&& a[i] == 9 && a[i + 1] == 4)
vis = 1;
}
return ans + vis;
}
int main()
{
int t;
f();
/*int i;
for(i = 1 ;i <=18; i++)
cout<<dp[i][0]<<" "<<dp[i][1]<<" "<<dp[i][2]<<endl;*/
cin>>t;
while(t--)
{
cin>>n;
LL ans = ff();
cout<<ans<<endl;
}
return 0;
}