codeforces587B Duff in Beach

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题意就是两个数组A，B其中B[i] = A[i%n]，B有l个元素，A有n个元素。

从B中选取x个元素，相邻元素在B中的下标满足Ij/n + 1 == I[j+1]/n。

可以用dp[i][j]表示第i个数选的是j，那么dp[i][j] = ∑dp[i-1][t] && t <= j。

数字比较大，得先离散化一下，且要用滚动数组。。。。

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2016File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtypedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;typedef pair<ii,int> iii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;LL dp[2][1000010];LL a[1000010],b[1000010];int n, k;LL l;int main(){        // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    cin >> n >> l >> k;    for (int i = 1;i <= n;++i) {        // cin >> a[i];        scanf("%I64d",&a[i]);        b[i] = a[i];    }    sort(b + 1,b + 1 + n);    int cnt = unique(b + 1,b + 1 + n) - b;    for (int i = 1;i <= n;++i) {        a[i] = lower_bound(b + 1,b + cnt, a[i]) - b;        // cout << a[i] << ' ';    }    // cout << endl;    int now = 1, la = 0;    dp[now][0] = 1;    LL ans = 0;    LL seg = l / n;    for (int i = 1;i <= k && i <= seg + 1;++i) {        swap(la, now);        for (int j = 0;j <= cnt;++j)            dp[now][j] = 0;        for (int j = 1;j <= cnt;++j)            dp[la][j] = (dp[la][j-1] + dp[la][j]) % MOD;        for (int j = 1;j <= n;++j)            dp[now][a[j]] = (dp[now][a[j]] + dp[la][a[j]]) % MOD;        for (int j = 1;j <= cnt;++j)            ans = (ans + (seg - i + 1)%MOD * dp[now][j]%MOD) % MOD;        for (int j = 1;j <= l - seg*n;++j)            ans = (ans + dp[la][a[j]]) % MOD;    }    cout << ans << endl;    return 0;}

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