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 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.
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).
Input
3 2
1 3
2 3
4 4
Output
6
5
5
For K = 2 and length 1 Marmot can eat ®.
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).
看了好久的题目,才明白,k朵白花为一组,吃白花必须一组一起吃,可以吃好几组,提示里面的R和W分别是 Red红 和 White白 。
dp动态规划题,我是直接写了份暴力的代码找规律,发现在k之前全都是1,后面大于k时,我们dp[i]可以是dp[i - 1]加上一朵红花,或者dp[i - k]加上k朵白花,dp[i] = dp[i - 1] + dp[i - k] ,最后还需要求出每一次的和,然后往回减,否则复杂度高还是TLE。
#include
using namespace std;
#define INF 0x3f3f3f3f
const int maxn = 1e5+5;
typedef long long ll;
const ll mod = 1e9+7;
ll dp[maxn];
ll sum[maxn];
int main() {
ll t,k;
ll a,b;
cin >> t >> k;
for(int i=0;i<k;i++) {
dp[i] = 1;
}
for(int i=k;i<=maxn;i++) {
dp[i] = dp[i-1] + dp[i-k];
dp[i] %= mod;
}
sum[0] = 0;
for(int i=1;i<maxn;i++) {
sum[i] = sum[i-1] + dp[i];
sum[i] %= mod;
}
for(ll i=0;i<t;i++) {
cin >> a >> b;
cout << (sum[b] - sum[a-1] + mod) % mod << endl;
}
}
CodeForces 474D Flowers