hdu 5898 odd-even number (数位dp)

odd-even number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 454    Accepted Submission(s): 245

Problem Description

For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).

 

 

Input

First line a t,then t cases.every line contains two integers L and R.

 

 

Output

Print the output for each case on one line in the format as shown below.

 

 

Sample Input

2

1 100

110 220

 

 

Sample Output

Case #1: 29

Case #2: 36

 

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int dig[20];
ll dp[20][20][10];

ll dfs(int pos, int len, int oi, int flag0, int lim) {
    if(pos == -1) return ((len % 2 == 0 && oi % 2 != 0) || (len % 2 != 0 && oi % 2 == 0));
    if(!lim && dp[pos][len][oi] != -1) return dp[pos][len][oi];
    int End = lim ? dig[pos] : 9;
    ll ret = 0;
    for(int i = 0; i <= End; i++) {
        if(i == 0 && flag0) ret += dfs(pos - 1, 0, 0, 1, (i==End) && lim);
        else {
            int nlen, noi;
            noi = i;
            if(noi % 2 == oi % 2) nlen = len + 1;
            else {
                if(!((len % 2 == 0 && oi % 2 != 0) || (len % 2 != 0 && oi % 2 == 0) || len == 0)) {
                    continue;
                }
                else nlen = 1;
            }
            ret += dfs(pos - 1, nlen, noi, 0, (i==End) && lim);
        }
    }
    if(!lim) dp[pos][len][oi] = ret;
    return ret;
}

ll func(ll num) {
    int n = 0;
    while(num) {
        dig[n++] = num % 10;
        num /= 10;
    }
    return dfs(n - 1, 0, 0, 1, 1);
}

int main() {
    int t;
    scanf("%d", &t);
    int cas = 0;
    memset(dp, -1, sizeof(dp));
    while(t--) {
        ll l, r;
        scanf("%I64d %I64d", &l, &r);
        printf("Case #%d: %I64d\n", ++cas, func(r) - func(l-1));
    }
}