Attack on Titans（递推dp 条件限制）

Attack on Titans

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Over centuries ago, mankind faced a new enemy, the Titans. The difference of power between mankind and their newfound enemy was overwhelming. Soon, mankind was driven to the brink of extinction. Luckily, the surviving humans managed to build three walls: Wall Maria, Wall Rose and Wall Sina. Owing to the protection of the walls, they lived in peace for more than one hundred years.

But not for long, a colossal Titan appeared out of nowhere. Instantly, the walls were shattered, along with the illusory peace of everyday life. Wall Maria was abandoned and human activity was pushed back to Wall Rose. Then mankind began to realize, hiding behind the walls equaled to death and they should manage an attack on the Titans.

So, Captain Levi, the strongest ever human being, was ordered to set up a special operation squad of N people, numbered from 1 to N. Each number should be assigned to a soldier. There are three corps that the soldiers come from: the Garrison, the Recon Corp and the Military Police. While members of the Garrison are stationed at the walls and defend the cities, the Recon Corps put their lives on the line and fight the Titans in their own territory. And Military Police serve the King by controlling the crowds and protecting order. In order to make the team more powerful, Levi will take advantage of the differences between the corps and some conditions must be met.

The Garrisons are good at team work, so Levi wants there to be at least M Garrison members assigned with continuous numbers. On the other hand, members of the Recon Corp are all elite forces of mankind. There should be no more than K Recon Corp members assigned with continuous numbers, which is redundant. Assume there is unlimited amount of members in each corp, Levi wants to know how many ways there are to arrange the special operation squad.

Input

There are multiple test cases. For each case, there is a line containing 3 integers N (0 < N < 1000000), M (0 < M < 10000) and K (0 < K < 10000), separated by spaces.

Output

One line for each case, you should output the number of ways mod 1000000007.

Sample Input

3 2 2

Sample Output

5

Hint

Denote the Garrison, the Recon Corp and the Military Police as G, R and P. Reasonable arrangements are: GGG, GGR, GGP, RGG, PGG.

题意：有三个兵团 GRP，现在需要组成N人团队，至少调动M个连续G士兵，至多调动K个连续R士兵，问有多少种组合情况

思路：由于又是至多又是至少没办法处理，所以统一转换成至多，对于事件A={至多N个G士兵连续}，事件B={至多m-1个G士兵连续},那么A-B={连续人数在m到n中取值，即至少m个连续士兵}。统一之后，设定一个二维数组，dp[i][3]，dp[i][0]表示对于第i个位置，放G士兵，同理,dp[i][1],dp[i][2]分别表示第i个位置放R，和P。

         最大连续u个G，连续v个R；

        对于当前第i个位置，sum=dp[i-1][0]+dp[i-1][1]+dp[i-1][2];

        由于放P对整个没有影响，所以，dp[i][2]=sum;

        对于A事件  

                       当i <= u 时,dp[i][0]=sum;                                             //至多连续u个，那么小于等于u，就可以直接放G

                       当i==u+1时，dp[i][0]=sum-1;                                      // 对于现在这个位置，之前可能出现连续u个的情况，当前要放G，就要减去那一种情况

                       当i>u+1时，dp[i][0]=sum-dp[i-u-1][1]-dp[i-u-1][2];      //此时要减去i前面已经出现连续u个G的情况，即从i-u到i-1这一段都是G，那么i-1-u的位置可以是P或者R，                                                                                                          //于是减去dp[i-u-1][1]，dp[i-1-u][2]合起来总共具有的情况数。

         对于B事件                                                                                 //对于B事件同理

                         当i<=v时，dp[i][1]=sum;

                         当i==v+1时，dp[i][1]=sum-1;

                         当i>v+1时，dp[i][2]=sum-dp[i-v-1][0]-dp[i-v-1][2];

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Author: ZOU, Xu

#include <stdio.h>#include <string.h>const int mod = 1000000007;const int maxn = 1100000;long long dp[maxn][3];long long n, m, k;long long slove( long long u, long long v){   dp[0][0] = 1;   dp[0][1] = dp[0][2] = 0;   for( int i = 1; i <= n; i++)   {       long long sum = (dp[i-1][0] + dp[i-1][1] + dp[i-1][2]) % mod;       dp[i][2] = sum;       if( i <= u) dp[i][0] = sum;       if( i == u + 1) dp[i][0] = sum - 1;       if( i > u + 1)  dp[i][0] = (sum - dp[i - u - 1][1] - dp[i - u - 1][2]) % mod;       if( i <= v) dp[i][1] = sum;       if( i == v + 1) dp[i][1] = sum - 1;       if( i > v + 1)  dp[i][1] = (sum - dp[i - v - 1][2] - dp[i - v - 1][0]) % mod;      //printf("%d  %d  %d--\n",dp[i][0],dp[i][1],dp[i][2]);   }   return (dp[n][0] + dp[n][1] + dp[n][2]) % mod;}int main(){    while(~scanf("%lld%lld%lld",&n,&m,&k))    {         printf("%lld\n",((slove(n, k) - slove(m-1, k))%mod + mod)%mod);    }}

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