Codeforces Round #326 (Div. 2) D. Duff in Beach（LIS）

题意:

给定一个长度为N的序列a,给定一个长度为L≤1018的序列b,且bi=ai%N
求长度不超过K的b的不降子序列的个数,这个LIS的每个元素不能在同一个N周期内

分析:

由于每个周期最多只能选择一个,将a序列排序
考虑dp[i][j]:=长度为i,且以有序序列a第j个结尾的LIS的个数
转移的时候我们可以维护一个指针,这样转移就是均摊O(1)的了
统计答案时,由于是连续周期的,假如可选周期是T,当前长度为i,那么应该乘上(T−i+1)

代码:

////  Created by TaoSama on 2016-01-26//  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 = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n, K; LL l;pair<int, int> a[N];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);    while(scanf("%d%I64d%d", &n, &l, &K) == 3) {        for(int i = 0; i < n; ++i) {            scanf("%d", &a[i].first);            a[i].second = i;        }        sort(a, a + n);        vector<vector<int> > f(K, vector<int>(n));        for(int i = 0; i < n; ++i) f[0][i] = 1;        for(int i = 1; i < K; ++i) {            int sum = 0;            for(int j = 0, k = 0; j < n; ++j) {                while(k < n && a[k].first <= a[j].first)                    sum = (sum + f[i - 1][k++]) % MOD;                f[i][j] = sum;            }        }        LL ans = 0;        for(int i = 0; i < K; ++i) {            for(int j = 0; j < n; ++j) {                LL cnt = l / n + (a[j].second < l % n);                if(cnt - i > 0) {                    ans += (cnt - i) % MOD * f[i][j] % MOD;                    ans %= MOD;                }            }        }        printf("%I64d\n", ans);    }    return 0;}