UVa10943

10943 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 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
21

题意：

       将K个不超过N的非负整数加起来，使得它们的和为N，有多少种方法？N=5，K=2时一共有6种方法，即0+5、1+4、2+3、3+2、4+1、5+0。输出方法总数模1000000的余数。

分析：

       相当于解方程sum{xi | i = 1,2,…,K && xi >= 0}。答案就是C(N+K-1,K-1)。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define ll long long
const int MOD = 1000000;
const int maxk = 200;
ll C[maxk + 2][maxk + 2];
// 线性算法，可以加取模
void get_C(){
    memset(C,0,sizeof C);
    C[0][0] = 1;
    for(int i = 0 ; i <= maxk ; i++){
        C[i][0] = C[i][i] = 1;
        for(int j = 1 ; j < i ; j++)
            C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
    }
}
// 直接计算，不要随便取模，计算量过大时会有误差
long long cal_C(long long n,long long m){
    double ans = 1;
    for(int i = 0 ; i < m ; i++) ans *= n - i;
    for(int i = 0 ; i < m ; i++) ans /= i + 1;
    return (long long)(ans + 0.5) % MOD;
}
int main(){
    int N,K;
    get_C();
    while(scanf("%d%d",&N,&K) == 2 && N){
        printf("%lld
",C[N + K - 1][K - 1]);
    }
    return 0;
}