dp[i][j] 表示 用 前i种前能 凑成j 元的方法数。
dp[i][j] = dp[i-1][j] + dp[i][j- (i+1)] 因为我i是从0开始的。。 所以i+1 表示第i中的值
最后要用一下高精度
题目：
Appoint
description:
Description
Farmer John goes to Dollar Days at The Cow Store
and discovers an unlimited number of tools on sale. During his first visit, the
tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to
spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at
$2. Of course, there are other combinations for a total of 5 different ways FJ
can spend all his money on tools. Here they are:
1 @ US$3 + 1 @ US$2Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$K (1 <= K <= 100).
1 @ US$3 + 2 @ US$1
1 @ US$2 + 3 @ US$1
2 @ US$2 + 1 @ US$1
5 @ US$1
Input
A single line with two space-separated integers: N
and K.
Output
A single line with a single integer that is the
number of unique ways FJ can spend his money.
Sample Input
5 3
Sample Output
5
代码：
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 int N,K; 6 int dp[1000+10][1000+10]; 7 8 void epl(int a,int b) 9 { 10 for(int i=0;i<=60;i++) 11 { 12 dp[a][i] = dp[b][i]; 13 } 14 } 15 16 void add(int a,int b) 17 { 18 for(int i=0;i<60;i++) 19 { 20 dp[a][i]+= dp[b][i]; 21 if(dp[a][i]>=10) 22 { 23 dp[a][i]%=10; 24 dp[a][i+1] ++; 25 } 26 } 27 } 28 29 int main() 30 { 31 cin>>K>>N; 32 dp[0][0]=1; 33 for(int i=0;i<N;i++) 34 { 35 for(int k=0;k<=K;k++) 36 { 37 //dp[i+1][k] = dp[i][k]; 38 if( k>=i+1) 39 { 40 //dp[i+1][k]+= dp[i+1][k-i-1]; 41 add(k,k-i-1); 42 } 43 } 44 } 45 int t = 60; 46 while(t>0 && dp[K][t]==0)t--; 47 for(int i=t;i>=0;i--) 48 { 49 printf("%d",dp[K][i]); 50 } 51 //cout<<dp[N][K]; 52 printf("\n"); 53 return 0; 54 }