#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n, r, t, i, ans, cs=1, up, dwn, fac[1000009];
const ll mod=1000003;
ll bigmod(ll b, ll p){
if(p==0) return 1;
ll x=bigmod(b, p/2);
x=(x*x)%mod;
if(p%2==1)x=(x*b)%mod;
return x;
}
int main(){
fac[0]=1;
for(i=1; i<=1000000; i++){
fac[i]=(fac[i-1]*i)%mod;
}
cin>>t;
while(t--){
cin>>n>>r;
up=fac[n];
dwn=(fac[n-r]*fac[r])%mod;
ans=up*bigmod(dwn, mod-2);
cout<<"Case "<<cs++<<": "<<ans%mod<<endl;
}
return 0;
}
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Friday, January 20, 2017
Solution of Light OJ 1067-Combinations
See the problem Light OJ 1067
Solution of Light OJ 1028 -Trailing Zeroes (I)
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll t, n, m, i, j, ans, prime[1000009], num[1000009]={0}, cnt, cs=1;
void seive(){
for(i=3; i*i<=1000000; i+=2){
if(num[i]==0){
for(j=i*i; j<=1000000; j=j+(i*2)) num[j]=1;
}
}
prime[0]=2;
j=1;
for(i=3; i<=1000000; i+=2){
if(num[i]==0) prime[j++]=i;
}
}
int main(){
seive();
scanf("%lld", &t);
while(t--){
cin>>n;
ans=1;
for(i=0; i<j&&prime[i]*prime[i]<=n; i++){
cnt=0;
while(n%prime[i]==0){
cnt++;
n=n/prime[i];
}
ans=ans*(cnt+1);
}
if(n!=1)ans=ans*2;
cout<<"Case "<<cs++<<": "<<ans-1<<endl;
}
return 0;
}
Thursday, January 19, 2017
Solution of UVa-11029 Leading and Trailing
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n, k, t_case;
ll bigmod(ll b, ll p, ll m){
if(p==0)return 1;
ll xx=bigmod(b, p/2, 1000);
xx=(xx*xx)%1000;
if(p%2==1)xx=(xx*b)%1000;
return xx;
}
int main(){
cin>>t_case;
while(t_case){
cin>>n>>k;
/* executing first 3 digits */
double x=k*(log10(n));
x=x-(int)x; // taking fraction value only
double ans=pow(10, x);
ans=ans*100;
cout<<(int)ans<<"...";
/* executing last 3 digits */
printf("%03d\n", bigmod(n, k, 1000));
t_case--;
}
return 0;
}
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