10943 How do you add?【組合數取餘（遞推）】
阿新 • • 發佈：2019-02-10
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,
20 2
20 2
0 0 Sample Output
21
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.
20 2
20 2
0 0 Sample Output
21
21
求C(n,m)%10^6;
AC程式碼:
#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; }