[HDU3555]Bomb

[HDU3555]Bomb

试题描述

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?

输入

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.

输出

For each test case, output an integer indicating the final points of the power.

输入示例

3
1
50
500

输出示例

0
1
15

数据规模及约定

见“输入”

题解

裸数位 dp，f[i][j] 表示前 i 位最高位为数字 j 的数中，包含“49”的个数。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
#define LL long long

LL read() {
    LL x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

#define maxn 21
LL f[maxn][maxn], ten[maxn];

int num[maxn], cnt;
LL sum(LL x) {
	cnt = 0; LL orx = x;
	while(x) num[++cnt] = x % 10, x /= 10;
	LL sum = 0;
	for(int i = cnt; i; i--) {
		for(int j = 0; j < num[i]; j++) sum += f[i][j];
		if(i < cnt && num[i+1] == 4 && num[i] == 9) {
			sum += orx % ten[i-1] + 1; break;
		}
	}
	return sum;
}

int main() {
	ten[0] = 1;
	for(int i = 1; i < maxn; i++) ten[i] = ten[i-1] * 10;
	for(int i = 1; i < maxn - 1; i++)
		for(int j = 0; j <= 9; j++)
			for(int k = 0; k <= 9; k++) {
				if(k == 4 && j == 9) f[i+1][k] += ten[i-1];
				else f[i+1][k] += f[i][j];
			}
	
	int T = read();
	while(T--) {
		LL n = read();
		printf("%lld\n", sum(n));
	}
	
	return 0;
}

打这种题最好写一发暴力。。。