UVA - 10943 How do you add?【组合数取余（递推）】

 Larry is very bad at math — he usually uses a calculator, whichworked well throughout college. Unforunately, he is now struck ina deserted island with his good buddy Ryan after a snowboardingaccident.They’re now trying to spend
some time figuring out some goodproblems, and Ryan will eat Larry if he cannot answer, so his fateis up to you!It’s a very simple problem — given a number N, how many wayscan K numbers less than N add up to N?For example, for N = 20 and K = 2, there are 21
ways:0+201+192+183+174+165+15...18+219+120+0InputEach 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.OutputSince 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 Input20 220 20 0Sample Output2121
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

求C(n,m)%10^6;

 

AC代码:

#include<cstdio>
#include<cstring>

typedef long long LL;
const LL MOD=1e6;

LL a[333][333];

int main()
{
LL N,K;
memset(a,0,sizeof(a));
for(int i=0;i<222;++i) a[i][0]=1;
for(int i=1;i<222;++i) {
for(int j=1;j<=i;++j) a[i][j]=(a[i-1][j]+a[i-1][j-1])%MOD;
}
while(~scanf("%lld%lld",&N,&K),N||K) {
printf("%lld\n",a[N+K-1][K-1]);
}
return 0;
}