2014
11-05

# Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7487    Accepted Submission(s): 2610

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.  题意：给出n，求出[1，n]中多少个数包含49；思路：数位dp。1.dp[len][0] 代表数字长度为len不含49的个数 2.dp[len][1] 代表数字长度为len不含49但是以9开头的个数（显然dp[len][1]包含在dp[len][0]中）3.dp[len][2] 代表数字长度为len含有49的个数 AC代码：#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define ll long long
using namespace std;

const int maxn = 3005;
const int INF = 1e9;

ll dp[25][3];
ll digit[25];
void init(){
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for(int i = 1; i < 21; i++)
{
dp[i][0] = 10 * dp[i - 1][0] - dp[i - 1][1];   //在前面添上0~9，要减去4加到9前面这种情况
dp[i][1] = dp[i - 1][0];
dp[i][2] = 10 * dp[i - 1][2] + dp[i - 1][1];
}
}
int main()
{
init();
int t;
ll n;
scanf("%d", &t);
while(t--)
{
scanf("%I64d", &n);
int len = 1;
memset(digit, 0, sizeof(digit));
while(n)
{
digit[len++] = n % 10;
n /= 10;
}
ll ans = 0;
bool flag = 0;
for(int i = len - 1; i >= 1; i--)
{
ans += dp[i - 1][2] * digit[i];
if(flag) ans += digit[i] * dp[i - 1][0];
else if(digit[i] > 4) ans += dp[i - 1][1];
if(digit[i + 1] == 4 && digit[i] == 9) flag = 1;
}
if(flag) ans++;
printf("%I64d\n", ans);
}
return 0;
}



1. 站长，你好！
你创办的的网站非常好，为我们学习算法练习编程提供了一个很好的平台，我想给你提个小建议，就是要能把每道题目的难度标出来就好了，这样我们学习起来会有一个循序渐进的过程！

2. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧