D - Combinations
Given n different objects, you want to take k of them. How many ways to can do it?
For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways.
Take 1, 2
Take 1, 3
Take 1, 4
Take 2, 3
Take 2, 4
Take 3, 4
Input
Input starts with an integer T (≤ 2000), denoting the number of test cases.
Each test case contains two integers n (1 ≤ n ≤ 106), k (0 ≤ k ≤ n).
Output
For each case, output the case number and the desired value. Since the result can be very large, you have to print the result modulo 1000003.
Sample Input
3
4 2
5 0
6 4
Sample Output
Case 1: 6
Case 2: 1
Case 3: 15
这题目有两种方法
第一种是通俗易懂但要复杂一点的方法，就是你要分解质因数，用桶排记录它们的个数，组合的公式 C（n，m）= n！/（m！*（n-m）！）将n! 的所有质因数累加，m！和（n-m)！的质因数减去，
最后算出这些质因数相乘的值。这就要用到线性筛啊，快速幂等等
附上代码（很长)
1 var 2 n,m,i,j,k,t,sum,p:longint; 3 ans:int64; 4 mark:array[1..200000] of boolean; 5 prim,num,markx:array[1..200000] of int64; 6 function prime(x:longint):longint; 7 var i,j,k:longint; 8 begin 9 k:=0; 10 for i:=2 to x do 11 begin 12 if mark[i] 13 then 14 begin 15 k:=k+1; 16 prim[k]:=i; 17 markx[i]:=k; 18 end; 19 for j:=1 to k do 20 begin 21 if i*prim[j]>m 22 then break; 23 mark[i*prim[j]]:=false; 24 if i mod prim[j]=0 25 then break; 26 end; 27 end; 28 prime:=k; 29 end; 30 function f(a,b:longint):int64; 31 var y:int64; 32 begin 33 t:=1; 34 y:=a; 35 while b<>0 do 36 begin 37 if (b and 1)=1 38 then t:=t*y mod p; 39 y:=y*y mod p; 40 b:=b shr 1; 41 end; 42 f:=t; 43 end; 44 45 begin 46 readln(m,n,p); 47 for i:=1 to m do 48 mark[i]:=true; 49 sum:=prime(m); 50 for i:=2 to m do 51 begin 52 if mark[i] 53 then 54 begin 55 inc(num[markx[i]]); 56 continue; 57 end; 58 k:=i; 59 j:=1; 60 while (prim[j]<=trunc(sqrt(i)))and(k>1) do 61 begin 62 if k mod prim[j]=0 63 then 64 while k mod prim[j]=0 do 65 begin 66 k:=k div prim[j]; 67 num[j]:=num[j]+1; 68 end; 69 j:=j+1; 70 end; 71 if k>1 72 then inc(num[markx[k]]); 73 end; 74 for i:=2 to n do 75 begin 76 if mark[i] 77 then 78 begin 79 dec(num[markx[i]]); 80 continue; 81 end; 82 k:=i; 83 j:=1; 84 while (prim[j]<=trunc(sqrt(i)))and(k>1) do 85 begin 86 if k mod prim[j]=0 87 then 88 while k mod prim[j]=0 do 89 begin 90 k:=k div prim[j]; 91 num[j]:=num[j]-1; 92 end; 93 j:=j+1; 94 end; 95 if k>1 96 then dec(num[markx[k]]); 97 end; 98 for i:=2 to m-n do 99 begin 100 if mark[i] 101 then 102 begin 103 dec(num[markx[i]]); 104 continue; 105 end; 106 k:=i; 107 j:=1; 108 while (prim[j]<=trunc(sqrt(i)))and(k>1) do 109 begin 110 if k mod prim[j]=0 111 then 112 while k mod prim[j]=0 do 113 begin 114 k:=k div prim[j]; 115 num[j]:=num[j]-1; 116 end; 117 j:=j+1; 118 end; 119 if k>1 120 then dec(num[markx[k]]); 121 end; 122 123 ans:=1; 124 for i:=1 to sum do 125 if num[i]>0 126 then ans:=(ans*f(prim[i],num[i])) mod p; 127 writeln(ans); 128 end.
第二种就要简单多了，直接算，但是因为要取模，模数又很大，不能直接取模，要用到逆元，就是乘上逆元再模。求逆元呢，要用到扩展欧几里得。再就是一点组合数的运算了
1 var 2 l,t,i,modd:longint; 3 f,inv:array[0..1100005]of int64; 4 n,m:int64; 5 function exgcd(a,b:int64;var x,y:int64):int64; 6 var r,t:int64; 7 begin 8 if b=0 then 9 begin 10 x:=1; 11 y:=0; 12 exit(a); 13 end; 14 r:=exgcd(b,a mod b,x,y); 15 t:=x; 16 x:=y; 17 y:=t-a div b*y; 18 exit(r); 19 end; 20 function cal(d,m:int64):int64; 21 var x,y:int64; 22 begin 23 exgcd(d,m,x,y); 24 exit((x mod m+m) mod m); 25 end; 26 function c(a,b:int64):int64; 27 begin 28 exit(f[a]*inv[b]mod modd *inv[a-b] mod modd); 29 end; 30 begin 31 readln(t); 32 modd:=1000003; 33 f[0]:=1;inv[0]:=1; 34 for i:=1 to 1010000 do 35 begin 36 f[i]:=f[i-1]*i mod modd; 37 inv[i]:=cal(f[i],modd); 38 end; 39 40 for l:=1 to t do 41 begin 42 readln(n,m); 43 // if m=0 then begin writeln(1); continue; end; 不要想当然乱加条件，先试试程序对于特殊样例能不能过，能过就不要加初始条件了 44 writeln('Case ',l,': ',c(n,m)); 45 end; 46 end.