Codeforces Round #271 (Div. 2) D 递推

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D. Flowers

time limit per test1.5 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of them white and some of them red.

But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.

Now Marmot wonders in how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo 1000000007 (109 + 7).

Input
Input contains several test cases.

The first line contains two integers t and k (1 ≤ t, k ≤ 105), where t represents the number of test cases.

The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105), describing the i-th test.

Output
Print t lines to the standard output. The i-th line should contain the number of ways in which Marmot can eat between ai and bi flowers at dinner modulo 1000000007 (109 + 7).

Examples
input
3 2
1 3
2 3
4 4
output
6
5
5
Note
For K = 2 and length 1 Marmot can eat (R).
For K = 2 and length 2 Marmot can eat (RR) and (WW).
For K = 2 and length 3 Marmot can eat (RRR), (RWW) and (WWR).
For K = 2 and length 4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR).

题意：

给出n,k分别表示花的朵数以及每一团白花的大小。

要求输出区间长度[a,b]内，白花的所有摆放种数。（这里一旦出现白花那么size必须是k）

思路：

手推几组样例就发现了长度为i且size为k的时候方案数为： a[i]=a[i-1]+a[i-k]

然后用个前缀和 sum[i]=sum[i-1]+a[i]

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#include<bitset>#define mst(ss,b) memset((ss),(b),sizeof(ss))///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;#define INF (1ll<<60)-1#define Max 1e9#define mod 1000000007using namespace std;ll sum[100100],a[100100];int main(){    int T,k;    scanf("%d%d",&T,&k);    sum[0]=0;    for(int i=1;i<k;i++) a[i]=1;    a[k]=2;    for(int i=k+1;i<=100000;i++){        a[i]=(a[i-1]+a[i-k])%mod;    }    for(int i=1;i<=100000;i++) sum[i]=(sum[i-1]+a[i])%mod;    while(T--){        int x,y;        scanf("%d%d",&x,&y);        printf("%I64d\n",((sum[y]-sum[x-1])%mod+mod)%mod);    }    return 0;}