UVA - 11038 How Many O's? (数位dp)

How Many O‘s?

题意是求区间内数字中0的个数，比如100就有两个0。

数位dp吧，dp[i][j][k], i很明显表示当前位置，j表示找到的0的个数，k表示要找的0的个数。因为数字里0的个数最多32个，所以可以枚举32种k的情况，用数位dp去找。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define ll long long
int dig[20];
ll dp[40][40][40];
int bigcnt;
ll dfs(int pos, int cnt, int flag, int lim) {
    if(pos == -1) return cnt == 0;
    if(cnt < 0) return 0;
    if(!flag && !lim && dp[pos][cnt][bigcnt] != -1) return dp[pos][cnt][bigcnt];
    int End = lim ? dig[pos] : 9;
    ll ret = 0;
    for(int i = 0; i <= End; i++) {
        if(flag && !i) ret += dfs(pos - 1, bigcnt, 1, lim && (i == End));
        else if(flag && i) ret += dfs(pos - 1, bigcnt, 0, lim && (i == End));
        else if(!flag && i) ret += dfs(pos - 1, cnt, 0, lim && (i == End));
        else if(!flag && !i) ret += dfs(pos - 1, cnt - 1, 0, lim && (i == End));
    }
    if(!lim && !flag) dp[pos][cnt][bigcnt] = ret;
    return ret;
}

ll func(ll num) {
    ll ret = 0;
    if(num == -1) return -1;
    int n = 0;
    while(num) {
        dig[n++] = num % 10;
        num /= 10;
    }
    for(int i = 1; i <= 32; i++) {
        bigcnt = i;
        ret += i * dfs(n - 1, i, 1, 1);
    }
    return ret;
}

int main() {
    ll a, b;
    memset(dp, -1, sizeof(dp));
    while(~scanf("%lld %lld", &a, &b)) {
        if(a < 0) break;
        printf("%lld\n", func(b) - func(a - 1));
    }
}

UVA - 11038 How Many O's? (数位dp)