HDU 4734 F(x) 数位dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4734

F(x)

Time Limit: 1000/500 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others) #### 问题描述 > For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A). #### 输入 > The first line has a number T (T <= 10000) , indicating the number of test cases. > For each test case, there are two numbers A and B (0 <= A,B < 109) #### 输出 > For every case,you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output the answer. ####样例输入 > 3 > 0 100 > 1 10 > 5 100 ####样例输出 > Case #1: 1 > Case #2: 2 > Case #3: 13

题解

数位dp，这题求的是<=的所有情况，这和求==的唯一差别就是初始化：return k>=0；

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef int LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=11;
const int maxm=4666;

int arr[maxn],tot;
int dp[maxn][maxm];
int bin[maxn];
///ismax标记表示前驱是否是边界值
LL dfs(int len,int k, bool ismax) {
    if(k<0) return 0;
    if (len == 0) {
        ///递归边界,求"恰好等于"和"小于等于"唯一的区别是：
        ///return k==0 VS return k>=0
        return k>=0;
    }
    if (!ismax&&dp[len][k]>=0) return dp[len][k];
    LL res = 0;
    int ed = ismax ? arr[len] : 9;

    ///这里插入递推公式
    for (int i = 0; i <= ed; i++) {
        res += dfs(len - 1, k-i*bin[len-1], ismax&&i == ed);
    }

    return ismax ? res : dp[len][k] = res;
}

LL solve(LL x,int y) {
    tot = 0;
    int k=0,tmp=1;
    while (x) {
        k+=(x % 10)*tmp; x /= 10;
        tmp*=2;
    }
    while (y) { arr[++tot] = y % 10; y /= 10; }
    return dfs(tot, k, true);
}

int main() {
    bin[0]=1;
    for(int i=1;i<maxn;i++) bin[i]=bin[i-1]*2;
    clr(dp,-1);
    int tc,kase=0;
    scf("%d",&tc);
    while(tc--){
        int x,y;
        scf("%d%d",&x,&y);
        prf("Case #%d: %d
",++kase,solve(x,y));
    }
    return 0;
}

//end-----------------------------------------------------------------------

Notes

一直在想求<=要怎么办，枚举（0,1,2,3，。。。，k）？ 复杂度有点高。。。。。
其实，把状态dp[len][k]定义成前len位所有<=k的情况，然后只要改下初始化就行了xrz