[Water + math] Fzu OJ 2193 so hard and 2191 perfect numbers

2193 so hard

Test instructions

Simplify decimals.

Ideas:

No more than 9 bits, then multiply 10^9 just fine.

Then note that the accuracy is added.

Code:

#include "cstdlib" #include "Cstdio" #include "CString" #include "Cmath" #include "queue" #include "algorithm" #include " iostream "#include" map "#define LL __int64#define eps 1e-11using namespace std;ll gcd (ll a,ll b) {    return b==0?a:gcd (b, A%B);} int main () {    int t;    cin>>t;    while (t--)    {        double x;        scanf ("%lf", &x);        ll fz= (LL) ((x+eps) *1000000000);        ll fm=1000000000;        ll TEP=GCD (FZ,FM);        printf ("%i64d/%i64d\n", Fz/tep,fm/tep);    }    return 0;}

2191 the perfect number

Test instructions

The perfect degree of each number within the [a, b] interval.

The so-called perfect degree is the number of a*a*b (0<A<=B) that x can break down into.

Ideas:

Within the [1,x] interval.

We enumerate A.

Because a<=b for each I, so to set up, the smallest number is I*i*i

However, we just have to judge how many i*i are in the range of i*i*i<=x.

For each i*i-i+1.

Then subtract.

Code:

#include "cstdlib" #include "Cstdio" #include "CString" #include "Cmath" #include "queue" #include "algorithm" #include " iostream "#include" map "#define LL __int64#define EPS 1e-11using namespace std;int main () {    ll x, y;    while (scanf ("%i64d%i64d", &x,&y)!=-1)    {        ll ansx=0,ansy=0;        x--;        for (ll i=1;i*i*i<=x;i++) ansx+=x/(i*i)-i+1;        for (ll i=1;i*i*i<=y;i++) ansy+=y/(i*i)-i+1;        printf ("%i64d\n", ANSY-ANSX);    }    return 0;}

[Water + math] Fzu OJ 2193 so hard and 2191 perfect numbers