玲珑杯 1091 - Black and White（dp计数）

DESCRIPTION

Constroy likes the game called Reversi. He has a long paper tape with n

grids, where each grid should fill by one black chess or one white chess exactly. Constroy dislikes the situation with a consecutive black chesses or b

consecutive white chesses, so he intends to know how many situaions satisfy his preference.

The answer may be so large, but you only need to give the answer modulo (109+7)

.

INPUT

The first line contains a positive integer T

, which represents there are T test cases. The following is test cases. For each test case: The only one line contains three integers a,b and n. It is guaranteed that no more than 50 test cases satisfy n≥104. 1≤T≤103,1≤a,b,n≤106

OUTPUT

For each test case, output in one line, contains one integer, which represents the number of situations satisfy his preference modulo (109+7)

.

SAMPLE INPUT

10

1 1 2

2 3 3

4 6 5

5 6 4

4 5 6

8 1 9

9 1 8

9 9 10

16 16 16

1000000 1000000 1000000

SAMPLE OUTPUT

0

4

29

16

53

0

1

1018

65534

235042057

题目大意：

一共有N个格子，对于这N个格子来讲，要么涂成颜色a，要么涂成颜色b，要求不能有连续的a个颜色a出现，也不能有连续的b个颜色b出现。

问有多少种分配方式。

解题思路：

统计计数问题，考虑dp，设定d[i][2]，其中：

①d[i][0]表示长度为i的格子，以a颜色结尾的情况数。

②d[i][1]表示长度为i的格子，以b颜色结尾的情况数。

则可以得到其状态转移方程：

d[i][0]=Σd[i-j][1]（0 < j < a）

d[i][1]=Σd[i-j][0] (0 < j < b)

还需要注意的是，如果每次循环累加和的话会超时，需要维护一个d[i][0]和d[i][1]的前缀和。

#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 1000005
#define mod 1000000007
int d[maxn][2],sum[maxn][2];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int a,b,n;
scanf("%d%d%d",&a,&b,&n);
memset(d,0,sizeof(d));
memset(sum,0,sizeof(sum));
d[0][0]=d[0][1]=1;
sum[0][0]=sum[0][1]=1;
for(int i=1;i<=n;i++)
{
if(i<a)
d[i][0]=(d[i][0]+sum[i-1][1])%mod;
else
d[i][0]=(sum[i-1][1]-sum[i-a][1]+mod)%mod;
if(i<b)
d[i][1]=(d[i][1]+sum[i-1][0])%mod;
else
d[i][1]=(sum[i-1][0]-sum[i-b][0]+mod)%mod;
sum[i][0]=(sum[i-1][0]+d[i][0])%mod;
sum[i][1]=(sum[i-1][1]+d[i][1])%mod;
}
int ans=(d
[0]+d
[1])%mod;
printf("%d\n",(ans+mod)%mod);
}
return 0;
}