LightOJ 1068 - Investigation （数位dp）

题意:

[l,r]区间,l,r∈[1,231)中本身能被k<104整除,且数位和也能被k整除的数的数目

分析:

乍一看以为空间炸了不可做,dp[i][k][k],但是发现−−数位和顶多就90不到,所以算的时候特判下就好了
dp[i][mod][sumdMod]:从高到低第i位,模k余数是mod,数位和模k余数是sumdMod的满足要求的数字数
dp[11][90][90]很随意就过了

代码:

////  Created by TaoSama on 2015-11-22//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int k, digit[15], dp[11][90][90];int dfs(int i, int m, int sm, int e) {    if(!i) return !m && !sm;    if(!e && ~dp[i][m][sm]) return dp[i][m][sm];    int to = e ? digit[i] : 9;    int ret = 0;    for(int d = 0; d <= to; ++d)        ret += dfs(i - 1, (m * 10 + d) % k, (sm + d) % k, e && d == to);    return e ? ret : dp[i][m][sm] = ret;}int calc(int x) {    int cnt = 0;    for(; x; x /= 10) digit[++cnt] = x % 10;    if(k > cnt * 9) return 1; //only 0 can    return dfs(cnt, 0, 0, 1);}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int kase = 0;    int t; scanf("%d", &t);    while(t--) {        int l, r; scanf("%d%d%d", &l, &r, &k);        memset(dp, -1, sizeof dp);        printf("Case %d: %d\n", ++kase, calc(r) - calc(l - 1));    }    return 0;}