codeforces 474D DP

 Flowers    CodeForces - 474D

题目原文：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).

题目大意：在长度ｌ在ａ和ｂ之间的字符串中找到连续ｋ个白色小花。问这种的字符串有多少种。

解题思路：这道题按照样例先写出　０　到　５　的最大种数。

发现得到的字符串分别是：

１　Ｒ　　没有也满足，因为满足０个ｋ连续。

２　ＲＲ　　ＷＷ

３　ＲＲＲ　ＷＷＲ　ＲＷＷ

４　ＲＲＲＲ　ＲＷＷＲ　　ＲＲＷＷ　ＷＷＲＲ　ＷＷＷＷ

５　ＲＲＲＲＲ　ＲＲＲＷＷ　ＲＲＷＷＲ　ＲＷＷＲＲ　　ＷＷＲＲＲ　ＷＷＷＷＲ　ＷＷＲＷＷ　ＲＷＷＷＷ

感觉使用ｄｐ做　然后估了估　分两种　最后第ｉ个位子取Ｒ　和最后第ｉ个位子取Ｗ

如果取Ｒ　那我前面ｉ－１的字符串是要满足最大　与我第ｉ个位子上的Ｒ无关

如果取Ｗ　那我前ｋ－１个字符必须取Ｗ　否则不满足题意　那我前ｉ－ｋ满足最大值即可

所以选择策略为ｄｐ【ｉ】＝ｄｐ【ｉ－ｋ】＋　ｄｐ【ｉ－１】

#include
#include
using namespace std;
const int MOD=1e9+7;
const int MAXN=1e5+5;
int dp[100005];
int sum[MAXN];
int k;
void preoperation()
{
    dp[0]=0;
//    cout <<'*'<< k <> t >> k;
    preoperation();
    for(int i=1;i<=t;i++)
    {
        int a,b;
        cin >> a >>b ;
        cout << ((sum[b]-sum[a-1])%MOD+MOD)%MOD<