【jzoj4906】【NOIP2016提高组】【†组合数问题】【数论】

题目大意

解题思路

可以用杨辉三角求解组合数，再用前缀和求解答案。因为我比较蠢，所以我就分解质因数，但是实际上只要对k取模就可以了。

code

#include<cmath>#include<cstring>#include<cstdio>#include<algorithm>#define LD double#define LL long long#define max(x,y) ((x>y)?x:y)#define min(x,y) ((x<y)?x:y)#define fo(i,j,k) for(int i=j;i<=k;i++)#define fd(i,j,k) for(int i=j;i>=k;i--)using namespace std;int const maxn=2000;int t,K,n,m,f[maxn+10][20],g[maxn+10][maxn+10],ss[9]={8,2,3,5,7,11,13,17,19},inf=1e9;int main(){    freopen("problem.in","r",stdin);    freopen("problem.out","w",stdout);    scanf("%d%d",&t,&K);    int ii;    fo(i,1,maxn){        ii=i;        fo(j,1,8){            f[i][j]=f[i-1][j];            while(ii%ss[j]==0){                f[i][j]++;                ii/=ss[j];            }        }    }    int kk,cnt,tmp;    fo(i,1,maxn){        fo(j,1,i){            g[i][j]=g[i-1][j]+g[i][j-1]-g[i-1][j-1];            kk=K;tmp=inf;            fo(k,1,8){                cnt=0;                while(kk%ss[k]==0){                    cnt++;                    kk/=ss[k];                }                if(cnt)tmp=min(tmp,(f[i][k]-f[j][k]-f[i-j][k])/cnt);            }            if((tmp!=inf)&&(tmp>0))g[i][j]++;        }        fo(j,i+1,maxn)            g[i][j]=g[i-1][j]+g[i][j-1]-g[i-1][j-1];    }    int n,m;    fo(i,1,t){        scanf("%d%d",&n,&m);        printf("%d\n",g[n][m]);    }    return 0;}