### Counting block combinations II题号：115 难度： 35 中英对照

NOTE: This is a more difficult version of Problem 114.

A row measuring n units in length has red blocks with a minimum length of m units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.

Let the fill-count function, F(m, n), represent the number of ways that a row can be filled.

For example, F(3, 29) = 673135 and F(3, 30) = 1089155.

That is, for m = 3, it can be seen that n = 30 is the smallest value for which the fill-count function first exceeds one million.

In the same way, for m = 10, it can be verified that F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value for which the fill-count function first exceeds one million.

For m = 50, find the least value of n for which the fill-count function first exceeds one million.

### Code


public final class p115 {
public static void main(String[] args) {
long start=System.nanoTime();
String result = run();
long end=System.nanoTime();
System.out.println(result);
System.out.println( (end-start)/1000000 + "ms" );
}

static public long dp[][]=new long[200][2];
static public String run(){
for(int i=0;i<50;i++){
dp[i][0]=1;
dp[i][1]=0;
}
dp[50][0]=1;//black
dp[50][1]=1;//red
for(int i=51;i<200;i++){
dp[i][0]=dp[i-1][0]+dp[i-1][1];
dp[i][1]=0;
for(int j=50;j<=i;j++)
dp[i][1]+=dp[i-j][0];
if(dp[i][0]+dp[i][1]>1000000){
return Integer.toString(i);
}
}
return null;
}

}
168
0ms