allCodes

所属分类:C/C++基础
开发工具:C++
文件大小:0KB
下载次数:0
上传日期:2024-02-03 17:22:29
上 传 者sh-1993
说明:  福辛尼
(fohinni)

文件列表:
BFS
BFS_final.cpp
LIS_O(N log K).cpp
Problem_A: Burhan.cpp
Uva 10013 - Super long sums
Uva 686 - Goldbach's Conjecture (II).cpp
at Rahat Vai.cpp
emu8086_MP_assignment_final_16Aug.zip
emu8086_StringReverseAndPalindromeCheck.asm
emu8086_subStringCheck.asm
srs_submit.zip

//# codeing //fohinni ......................................................................................................................... #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define sf scanf #define pf printf #define sfint(a) scanf("%d",&a) #define sfinttwo(a,b) scanf("%d %d",&a,&b) #define sfl(a,b) scanf("%ld %ld",&a,&b) #define sfll(a,b) scanf("%lld %lld",&a,&b) #define sfd(a,b) scanf("%lf %lf",&a,&b) #define sff(a,b) scanf("%f %f",&a,&b) #define lp1(i,n) for(i=0;i T setbit(T n, T pos){n=n|(1< T checkbit(T n, T pos){n=n&(1<>5]=setbit(prime[i>>5],i&31); for(i=3;i<=x;i+=2) { if(!checkbit(prime[i>>5],i&31)) { for(j=i*i;j<=n;j+=i) { prime[j>>5]=setbit(prime[j>>5],j&31); } } } for(i=2;i<=n;i++) { if(!checkbit(prime[i>>5],i&31)) { pr[plen++]=i; } } } mapmp; void divisor(int n) { int i,m=1,tmp,take=n; mp.clear(); for(i=0;i1) { tmp=powl(n,2); m*=((tmp-1) / (n-1)); } pf("Summation of the divisors of %d is %d :)\n",take,m); m=1; } int main() { int n; seieve(100); while(sfint(n)==1) { divisor(n); } return 0; }

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