Codeforces 1073 E - Segment Sum

E - Segment Sum

思路:

数位dp

我们平时做的数位dp都是求满足条件的数的个数, 这里要求满足条件的数的和

只要在原来的基础上求每一位的贡献就可以了,所以传参数时要传两个

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int MOD = ;
int k;
pll dp[][];
int a[], tot;
LL pw[];
pll dfs(int pos, int s, bool zero, bool limit) {
    if(!pos) return {__builtin_popcount(s) <= k, };
    if(!limit && !zero && ~dp[pos][s].fi) return dp[pos][s];
    int up = ;
    if(limit) up = a[pos];
    pll ans = {, };
    for (int i = ; i <= up; i++) {
        pll res;
        if(zero && i == ) res = dfs(pos-, s, zero, limit&&i==up);
        else res = dfs(pos-, s|(<<i), zero&&i==, limit&&i==up);
        (ans.fi = ans.fi + res.fi) %= MOD;
        (ans.se = ans.se + res.se + i*res.fi%MOD*pw[pos-]%MOD) %= MOD;
    }
    if(!limit && !zero) dp[pos][s] = ans;
    return ans;
}
void init() {
    pw[] = ;
    for (int i = ; i < ; i++) pw[i] = (pw[i-] * ) % MOD;
    for (int i = ; i < ; i++)
        for (int j = ; j < ; j++) dp[i][j].fi = dp[i][j].se = -;
}
LL solve(LL n) {
    tot = ;
    init();
    while(n) {
        a[++tot] = n % ;
        n /= ;
    }
    return dfs(tot, , , ).se;

}
int main() {
    LL l, r;
    scanf("%lld %lld %d", &l, &r, &k);
    printf("%lld\n", (solve(r) - solve(l-) + MOD) % MOD);
    return ;
}