题目大意：输入K和S，输出含有且仅有K个不超过S的具有不小于1的公因子的不同正整数的方案数(如果大于10000，输出10000)。
例如：input:3 10
output ：11 //{2,4,6},{2,4,8},{2,4,10},{2,6,8},{2,6,10},{2,8,10},{4,6,8},{4,6,10},{4,8,10},{6,8,10},{3,6,9}.
Time Limit:1000MS Memory
Limit:65536KB 64bit IO Format:%I64d
& %I64u
规模:2<=K<=S<=50。
理论基础：定理：C(m,n)=C(m-1,n)+C(m-1,n-1)(参见杨辉三角);
题目分析：
根据例子分析来看，就是如果：n/i=m(整数除法),那么以i为公因子的组合数的个数为：C(m,k)；通过数据规模分析我们知道用到的组合数C(m,n)中的m,n<=25。那么C(m,n)不需要上述的定理也可以有了，打表就OK了。有些人开始高兴了，那么是不是枚举所有大于1小于等于S/K的数，然后加起来就好呢？答案是否定的，为什么呢？因为：{m*n,m*n*2,m*n*3}既会出现在m,也会出现在n里。分析到这里，你一定已经会做了吧。接下来就是怎么解决重复的问题。
首先可以肯定的是重复时i=m*n一定是合数（这就不用多讲了吧），而且枚举的时候只需要枚举指数且最大的质数为23。那下来我们就只需再去掉i就可以了，暴力完破。
代码如下：
#include<iostream>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
typedef double db;
#define DBG 1
#define maa (1<<31)
#define mii ((1<<31)-1)
#define sl(c) ((int)(c).size()) //取字符串长度；
#define forl(i, a, b) for(int i = (a); i < (b); ++i) //不带边界值循环，正序
#define forle(i, a, b) for(int i = (a); i <= (b); ++i) //带边界值循环，正序
#define forh(i, a, b) for(int i = (a); i > (b); --i) //不带边界值，逆序
#define forhe(i, a, b) for(int i = (a); i >= (b); --i) //带边界值，逆序
#define forlc(i, a, b) for(int i##_b = (b), i = (a); i < i##_b; ++i) //带别名的循环，用于不可改变值
#define forlec(i, a, b) for(int i##_b = (b), i = (a); i <= i##_b; ++i)
#define forgc(i, a, b) for(int i##_b = (b), i = (a); i > i##_b; --i)
#define forgec(i, a, b) for(int i##_b = (b), i = (a); i >= i##_b; --i)
#define forall(i, v ) forl(i, 0, sz(v)) //循环所有
#define forallc(i, v ) forlc(i, 0, sz(v))
#define forlla(i, v ) forhe(i, sz(v)-1, 0)
#define forls(i, n, a, b) for(int i = a; i != b; i = n[i]) //搜表用
#define rep(n) for(int repp = 0; repp < (n); ++repp)
#define repc(n) for(int repp_b = (n), repp = 0; repp < repp_b; ++repp)
#define rst(a, v) memset(a, v, sizeof a) //把字符v填充到a reset 重置
#define cpy(a, b) memcpy(a, b, sizeof a) //copy b 的sizeof（a）个字符到a
#define rstn(a, v, n) memset(a, v, (n)*sizeof((a)[0])) //把字符v填充到a[n]之前的字节
#define cpyn(a, b, n) memcpy(a, b, (n)*sizeof((a)[0])) //copy b 的 n 个字符到a
#define ast(b) if(DBG && !(b)) { printf("%d!!|\n", __LINE__); while(1) getchar(); } //调试
#define dout DBG && cout << __LINE__ << ">>| "
#define pr(x) #x"=" << (x) << " | "
#define mk(x) DBG && cout << __LINE__ << "**| "#x << endl
#define pra(arr, a, b) if(DBG) {\
dout<<#arr"[] |" <<endl; \
forlec(i, a, b) cout<<"["<<i<<"]="<<arr[i]<<" |"<<((i-(a)+1)%8?" ":"\n"); \
if((b-a+1)%8) puts("");\
} //数列查看
#define rd(type, x) type x; cin >> x //读数
inline int rdi() { int d; scanf("%d", &d); return d; }
inline char rdc() { scanf(" "); return getchar(); }
inline string rds() { rd(string, s); return s; }
inline db rddb() { db d; scanf("%lf", &d); return d; }
template<class T> inline bool updateMin(T& a, T b) { return a>b? a=b, true: false; }
template<class T> inline bool updateMax(T& a, T b) { return a<b? a=b, true: false; }
int c[26][26]=
{
1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,3,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,4,6,4,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,5,10,10,5,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,6,15,20,15,6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,7,21,35,35,21,7,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,8,28,56,70,56,28,8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,9,36,84,126,126,84,36,9,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,10,45,120,210,252,210,120,45,10,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,11,55,165,330,462,462,330,165,55,11,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,12,66,220,495,792,924,792,495,220,66,12,1,0,0,0,0,0,0,0,0,0,0,0,0,0,\
1,13,78,286,715,1287,1716,1716,1287,715,286,78,13,1,0,0,0,0,0,0,0,0,0,0,0,0,\
1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1,0,0,0,0,0,0,0,0,0,0,0,\
1,15,105,455,1365,3003,5005,6435,6435,5005,3003,1365,455,105,15,1,0,0,0,0,0,0,0,0,0,0,\
1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,1820,560,120,16,1,0,0,0,0,0,0,0,0,0,\
1,17,136,680,2380,6188,12376,19448,24310,24310,19448,12376,6188,2380,680,136,17,1,0,0,0,0,0,0,0,0,\
1,18,153,816,3060,8568,18564,31824,43758,48620,43758,31824,18564,8568,3060,816,153,18,1,0,0,0,0,0,0,0,\
1,19,171,969,3876,11628,27132,50388,75582,92378,92378,75582,50388,27132,11628,3876,969,171,19,1,0,0,0,0,0,0,\
1,20,190,1140,4845,15504,38760,77520,125970,167960,184756,167960,125970,77520,38760,15504,4845,1140,190,20,1,0,0,0,0,0,\
1,21,210,1330,5985,20349,54264,116280,203490,293930,352716,352716,293930,203490,116280,54264,20349,5985,1330,210,21,1,0,0,0,0,\
1,22,231,1540,7315,26334,74613,170544,319770,497420,646646,705432,646646,497420,319770,170544,74613,26334,7315,1540,231,22,1,0,0,0,\
1,23,253,1771,8855,33649,100947,245157,490314,817190,1144066,1352078,1352078,1144066,817190,490314,245157,100947,33649,8855,1771,253,23,1,0,0,\
1,24,276,2024,10626,42504,134596,346104,735471,1307504,1961256,2496144,2704156,2496144,1961256,1307504,735471,346104,134596,42504,10626,2024,276,24,1,0,\
1,25,300,2300,12650,53130,177100,480700,1081575,2042975,3268760,4457400,5200300,5200300,4457400,3268760,2042975,1081575,480700,177100,53130,12650,2300,300,25,1\
};
int nprime[9]={2,3,5,7,11,13,17,19,23};
int main()
{
int k,s,cnt=0,temp=9;
scanf("%d%d",&k,&s);
forl(i,0,9)
{
if(s/nprime[i]<k)
{
temp=i;
break;
}
cnt+=c[s/nprime[i]][k];
}
forl(j,0,temp)
{
forl(i,j+1,temp)
{
if(s/(nprime[j]*nprime[i])<k)break;
cnt-=c[s/(nprime[j]*nprime[i])][k];
}
}
if(cnt<=10000)printf("%d\n",cnt);
else printf("10000\n");
return 0;
}
C(m,m)的生成代码如下：
c[0][0]={1};
for(int i=1;i<=m;i++)
{
for(int j=0;j<=i;j++)
{
if(i==j||j==0)c[i][j]=1;
else c[i][j]=c[i-1][j-1]+c[i-1][j];
}
}
by：Jsun-moon http://blog.csdn.net/Jsun_moon