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

題意：

?????? 將K個不超過N的非負整數加起來，使得它們的和為N，有多少種方法？N=5，K=2時一共有6種方法，即0+5、1+4、2+3、3+2、4+1、5+0。輸出方法總數模1000000的余數。

分析：

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

1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 using namespace std; 5 #define ll long long 6 const int MOD = 1000000; 7 const int maxk = 200; 8 ll C[maxk + 2][maxk + 2]; 9 // 線性算法，可以加取模 10 void get_C(){ 11 memset(C,0,sizeof C); 12 C[0][0] = 1; 13 for(int i = 0 ; i <= maxk ; i++){ 14 C[i][0] = C[i][i] = 1; 15 for(int j = 1 ; j < i ; j++) 16 C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD; 17 } 18 } 19 // 直接計算，不要隨便取模，計算量過大時會有誤差 20 long long cal_C(long long n,long long m){ 21 double ans = 1; 22 for(int i = 0 ; i < m ; i++) ans *= n - i; 23 for(int i = 0 ; i < m ; i++) ans /= i + 1; 24 return (long long)(ans + 0.5) % MOD; 25 } 26 int main(){ 27 int N,K; 28 get_C(); 29 while(scanf("%d%d",&N,&K) == 2 && N){ 30 printf("%lld\n",C[N + K - 1][K - 1]); 31 } 32 return 0; 33 } View Code

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轉載于:https://www.cnblogs.com/cyb123456/p/5837669.html

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