10943 - How do you add? UVA-优快云博客

本文链接：https://blog.youkuaiyun.com/abc123456789cbaabc/article/details/8652590

本文提供了解决数学问题的方法，即在一个虚拟的沙漠孤岛上，通过组合数学原理，计算小于给定数值N的K个数相加等于N的所有可能组合数量。该文详细解释了如何通过表格法来快速找到答案，并提供了AC过的代码实例，以帮助读者理解并解决类似问题。

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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

参考：http://hi.baidu.com/zyz913614263/item/c64c7b5e96d94b2094eb0561

本题是一个规律题！！！

注意必须在打表的过程中对每个元素取模否则答案将会是错误的！！！！

n k 1 2 3 4 5 …… 100
1    1 2 3 4 5…… 100
2    1 3 6 10 15……
3    1 4 10 20 35 ……
4    1 5 15 35 70 ……
规律为 a[i][j]=a[i][j-1]+a[i-1][j]数据量很小，打表就过了，注意对1000000取模
代码如下已ac：

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
long long nk[110][110];
int main()
{
    int n,k;
    int i ;
    for(i = 1;i<=100;++i)
       nk[1][i] = i;
    for(i = 1;i<=100;++i)
       nk[i][1] = 1;
    int j;
    for(i = 2;i<=100;++i)
      for(j = 2;j<=100;++j)
      nk[i][j] = (nk[i][j-1]+nk[i-1][j])%1000000;
    while(scanf("%d %d",&n,&k)==2)
    {
        if(n==0&&k==0)
        break;
        printf("%lld\n",nk[n][k]);
    }
    return 0;
}