HDOJ 3555 Bomb

、英雄互娱（杭州）

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 12282    Accepted Submission(s): 4407

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：与HDU2089一样的意思，要求1~n的范围内含有49的数字的个数
思路：按照2089的思路去做就好了

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;long long dp[25][3];void Init(){    memset(dp, 0, sizeof(dp));    dp[0][0] = 1;    int i;    for (i = 1;i <= 22;i++){        dp[i][0] = dp[i - 1][0] * 10 - dp[i - 1][1];        dp[i][1] = dp[i - 1][0];        dp[i][2] = dp[i - 1][2] * 10 + dp[i - 1][1];    }}long long solve(long long n){    long long i, item = n, len = 0, a[25], flag = 0, ans = 0;    while (n){        a[++len] = n % 10;        n /= 10;    }    a[len + 1] = 0;    for (i = len; i; i--){        ans += dp[i - 1][2] * a[i];        if (flag)            ans += dp[i - 1][0] * a[i];        if (!flag && a[i] > 4)            ans += dp[i - 1][1];        if (a[i + 1] == 4 && a[i] == 9)            flag = 1;    }    return ans;}int main(){    int t;    long long n;    scanf("%d", &t);    Init();    while (t--){        scanf("%I64d", &n);        printf("%I64d\n", solve(n + 1));    }    return 0;}