Problem A: How do you add?
Larry is very bad at math - he usually uses a calculator,
which worked well throughout college. Unforunately, he is now struck in a deserted island with his good buddy Ryan after a snowboarding accident. They're now trying to spend some time figuring out some good problems, and Ryan will eat Larry if he cannot answer,
so his fate is up to you!
It's a very simple problem - given a number N, how many ways can K numbers
less than Nadd up to N?
For example, for N = 20 and K = 2, there are 21 ways:
0+20
1+19
2+18
3+17
4+16
5+15
...
18+2
19+1
20+0
Input
Each line will contain a pair of numbers N and K. N and K will
both be an integer from 1 to 100, inclusive. The input will terminate on 2 0's.
Output
Since Larry is only interested in the last few digits of the answer, for each pair of numbers N and K,
print a single number mod 1,000,000 on a single line.
Sample Input
20 2 20 2 0 0
Sample Output
21 21
题目大意：N拆成K个正整数的和，有多少种方法？
dp[N][K] 记录和为N，由K个数组成
则有dp[N][K]=sum{ dp[N-1][k-i] }
#include <iostream>
using namespace std;
const int mod=1000000;
int dp[110][110];
void ini(){
dp[0][0]=1;
for(int k=1;k<=100;k++){
for(int sum=0;sum<=100;sum++){
for(int i=0;i<=100-sum;i++){
dp[sum+i][k]=(dp[sum+i][k]+dp[sum][k-1])%mod;
}
}
}
}
int main(){
ini();
int n,k;
while(cin>>n>>k && (n||k)){
cout<<dp[n][k]<<endl;
}
return 0;
}