UVA 题目10943 How do you add?（DP，整数拆分）

最新推荐文章于 2021-06-07 20:06:22 发布

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最新推荐文章于 2021-06-07 20:06:22 发布

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分类专栏：动态规划

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面对荒岛生存挑战，Larry需要解决一个数学难题以避免被好友Ryan吃掉。问题是如何用K个小于N的数字相加得到N，求解方案的数量。通过动态规划方法，给出了一段C++代码实现。

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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
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

题目大意：大小为n的数分成m组有多少种

ac代码

#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
#include<limits.h>
#define N 100010
#define mod 1000000
#define LL long long
using namespace std;
LL dp[110][110];
void DP()
{
    int i,j,k;
    for(i=0;i<=100;i++)
        dp[0][i]=1;
    for(i=1;i<=100;i++)
    {
        for(j=1;j<=100;j++)
        {
            for(k=0;k<=i;k++)
                dp[i][j]=(dp[i][j]+dp[k][j-1])%mod;
        }
    }
}
int main()
{
    int n,k;
    DP();
    while(scanf("%d%d",&n,&k)!=EOF,n||k)
    {
        printf("%lld\n",dp[n][k]);
    }
}