题目链接：点击打开链接

思路：

给定T k表示T组测试数据

每组case [l,r]

有2种物品a b，b物品必须k个连续出现

问摆成一排后物品长度在[l,r]之间的方法数 

思路：

dp[i] = dp[i-1]+dp[i-k];

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
using namespace std;
typedef long long ll;
template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
template <class T>
inline void pt(T x) {
    if (x < 0) {
        x = -x;
        putchar('-');
    }
    if(x>9) pt(x/10);
    putchar(x%10+'0');
}
/////////////////////////
const int N = 100000 + 2;
const int mod = 1000000007;
ll d[N], sum[N];
int main() {
    int cas, K, l, r;
    rd(cas); rd(K);
    d[0] = 1;
    for (int i = 1; i < N; ++i) {
        d[i] = d[i - 1];
        if (i >= K)
            d[i] += d[i - K];
        d[i] %= mod;
    }
    sum[0] = 1;
    for (int i = 1; i < N; ++i) {
        sum[i] = sum[i - 1] + d[i];
        sum[i] %= mod;
    }
    while (cas -- > 0) {
        rd(l); rd(r);
        pt(((sum[r]-sum[l-1]) % mod + mod) % mod);
        putchar('\n');
    }
    return 0;
}

Codeforces 474D Flowers dp（水