D. Flowers
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).
Sample test(s)
input
3 21 32 34 4
output
655
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： 有个人喜欢吃花， 有白花，红花两种花， 但是吃白色只能一次性吃K 朵或者不吃白花，现在问你吃n朵花的方案数是多少.
题解： dp[i]表示吃i朵的方案数，那么 dp[i+k]+=dp[i]用来更新 吃一次白花，dp[i+1]+=dp[i]用来更新吃一次红花
///1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define INF 0x7fffffff
inline ll read()
{
ll x=0,f=1;
char ch=getchar();
while(ch<‘0‘||ch>‘9‘)
{
if(ch==‘-‘)f=-1;
ch=getchar();
}
while(ch>=‘0‘&&ch<=‘9‘)
{
x=x*10+ch-‘0‘;
ch=getchar();
}
return x*f;
}
//****************************************
#define maxn 100000+5
#define mod 1000000007
int dp[maxn];
int main(){
int n=read(),k=read();
mem(dp);
dp[0]=1;
for(int i=0;i<=100000;i++){
if(i+1<=100000)
dp[i+1]=(dp[i+1]+dp[i])%mod;
if(i+k<=100000)
dp[i+k]=(dp[i+k]+dp[i])%mod;
}
ll sum[maxn];
mem(sum);
sum[0]=0;
for(int i=1;i<=100000;i++){
sum[i]+=sum[i-1]+dp[i];
}
int a,b;
for(int i=1;i<=n;i++){
scanf("%d%d",&a,&b);
cout<<(sum[b]-sum[a-1])%mod<<endl;
}
return 0;
}
代码