【数位DP】Bomb HDU - 3555

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最新推荐文章于 2024-03-28 17:14:18 发布

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分类专栏：知识体系数位DP 题意思考文章标签：数位DP

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本文介绍了一种使用数位动态规划（数位DP）解决特定类型问题的方法，并通过一个具体的例子进行了详细说明。该例题的目标是在1到N的范围内找出所有包含子序列“49”的数字，以帮助反恐精英计算炸弹爆炸的威力。

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Think:
1知识点：数位DP
2题意：询问区间[1, n]内有多少个数的数位中含有49(eg:49,149,249,1491等)

Bomb HDU - 3555——vjudge题目链接

 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

Sample Output

0
1
15

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.

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

int dig[24];/*数位分离*/
LL dp[24][2][14];

LL dfs(int pos/*当前数位*/, bool state/*状态是否满足*/, int pre/*前一位的值*/, bool is_max/*是否达到临界*/);
LL solve(LL n);/*求解区间[1, n]*/

int main(){
    memset(dp, -1, sizeof(dp));
    int T;
    LL n;
    scanf("%d", &T);
    while(T--){
        scanf("%lld", &n);
        printf("%lld\n", solve(n));
    }
    return 0;
}
LL solve(LL x){
    memset(dig, 0, sizeof(dig));
    int pos = 0;
    while(x){
        dig[pos++] = x%10;
        x /= 10;
    }
    return dfs(pos-1, 0, -1, 1);
}
LL dfs(int pos, bool state, int pre, bool is_max){
    if(pos == -1)
        return state;/*返回true表示当前状态满足条件进而累加，返回false表示当前状态不满足条件*/
    if(!is_max && dp[pos][state][pre] != -1)
        return dp[pos][state][pre];
    int up_top = 9;
    if(is_max)/*临界判断*/
        up_top = dig[pos];
    LL cnt = 0;
    for(int i = 0; i <= up_top; i++){
        if(pre == 4 && i == 9)/*条件满足*/
            cnt += dfs(pos-1, true, i, is_max && i == up_top);
        else
            cnt += dfs(pos-1, state, i, is_max && i == up_top);
    }
    if(!is_max)/*全局记忆化搜索:判断是否可记忆*/
        dp[pos][state][pre] = cnt;
    return cnt;/*返回当前状态的解*/
}