poj3181 Dollar Dayz

Description

Farmer John goes to Dollar days at the Cow Store and discovers a unlimited number of tools on sale. During his first visit, the tools is selling variously for $, $, and $. Farmer John has exactly $ to spend. He can buy 5 tools at $1 tool at $ and a additional 1 tool at $. Of course, there is other combinations for a total of 5 different ways FJ can spend all he money on tools. Here they is:

        1 @ us$3 + 1 @ us$2        1 @ us$3 + 2 @ us$1        1 @ us$2 + 3 @ us$1        2 @ us$2 + 1 @ us$1        5 @ us$1

Write a program than would compute the number of ways FJ can spend N dollars (1 <= n <=) at the Cow Store for OLS on sale with a cost of $ $K (1 <= K <= 100).

Input

A single line with the space-separated integers:n and K.

Output

A single, with a single integer, is the number of the unique ways FJ can spend.

Sample Input

5 3

Sample Output

5

Test instructions: Give you two number n,k, let you use 1 to k this k number represents n, ask there are several methods, the essence is the splitting of integers. Because the final result is larger than a long long, so with two long long concatenated, set two array a[][],b[][] to indicate that there is not more than a long long part and more than a long long.

where a[n][m] means n with some number of split, where the largest number of the program number is not more than M, drawing can be seen when n<m, A[i][j]=a[i][i]; when N>=m, a[i][j]= (a[i][j-1]+a[i-j][j))%inf; Initialize the time to make a[0][i]=1, because a[2][2]=a[0][2]+a[2][1];

6
5 + 1
4 + 2, 4 + 1 + 1
3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1
2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + 1

#include <stdio.h>
#include <string.h>
#define LL Long Long
ll a[1006][106],b[1006][106];
ll INF;
int main ()
{
int n,m,i,j;
inf=1;
for (i=1;i<=18;i++) {
inf*=10;
}
while (scanf ("%d%d", &n,&m)!=eof)
{
memset (A,0,sizeof (a));
Memset (b,0,sizeof (b));
for (i=1;i<=m;i++) {
A[0][i]=1;
}
for (i=1;i<=n;i++) {
A[i][1]=1;
}
for (j=1;j<=m;j++) {
A[1][j]=1;
}

for (j=2;j<=m;j++) {
for (i=2;i<=n;i++) {
if (i<j) {
A[i][j]=a[i][i];
B[i][j]=b[i][i];
}
else{
A[i][j]= (A[i][j-1]+a[i-j][j])%inf;
b[i][j]=b[i][j-1]+b[i-j][j]+ (A[i][j-1]+a[i-j][j])/inf;
}
printf ("%d%d%d\n", i,j,a[i][j]);
}
}
if (B[n][m])
printf ("%lld", B[n][m]);
printf ("%lld\n", A[n][m]);
}
return 0;
}

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poj3181 Dollar Dayz