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[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。。。
[HDU3555]Bomb