标签:splay RKE you sim not answer its 动态规划 转移
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 N add 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
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.
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.
20 2
20 2
0 0
21
21
将N分解成K个非负整数之和的方案数。
//////////////////////////////////////////////////////////////////////
//Target: UVa 10943 - How do you add?
//@Author: Pisceskkk
//Date: 2019-2-16
//////////////////////////////////////////////////////////////////////
#include<cstdio>
#define N 220
#define mod 1000000
#define ll long long
using namespace std;
int n,k,f[N][N];
int dfs(int a,int b){
if(a < b)return 0;
if(b == 0)return 1;
if(a == 0)return 1;
if(f[a][b])return f[a][b];
return f[a][b] = (dfs(a-1,b-1)+dfs(a-1,b))%mod;
}
int main(){
while(1){
scanf("%d %d",&n,&k);
if(!n && !k){
break;
}
printf("%d\n",dfs(n+k-1,n));
}
return 0;
}
标签:splay RKE you sim not answer its 动态规划 转移
原文地址:https://www.cnblogs.com/pisceskkk/p/10421427.html