uva 11260

極致的overflow題目，利用四角錐數再用BS找sqrt

//
//        GGGGGGGGGGGGG        CCCCCCCCCCCCC               AAA
//     GGG::::::::::::G     CCC::::::::::::C              A:::A
//   GG:::::::::::::::G   CC:::::::::::::::C             A:::::A
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C            A:::::::A
// G:::::G       GGGGGG C:::::C       CCCCCC           A:::::::::A
//G:::::G              C:::::C                        A:::::A:::::A
//G:::::G              C:::::C                       A:::::A A:::::A
//G:::::G    GGGGGGGGGGC:::::C                      A:::::A   A:::::A
//G:::::G    G::::::::GC:::::C                     A:::::A     A:::::A
//G:::::G    GGGGG::::GC:::::C                    A:::::AAAAAAAAA:::::A
//G:::::G        G::::GC:::::C                   A:::::::::::::::::::::A
// G:::::G       G::::G C:::::C       CCCCCC    A:::::AAAAAAAAAAAAA:::::A
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C   A:::::A             A:::::A
//   GG:::::::::::::::G   CC:::::::::::::::C  A:::::A               A:::::A
//     GGG::::::GGG:::G     CCC::::::::::::C A:::::A                 A:::::A
//        GGGGGG   GGGG        CCCCCCCCCCCCCAAAAAAA                   AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(__VA_ARGS__)
#define gettime() end_time=clock();printf("now running time is %.7f\n",(float)(end_time - start_time)/CLOCKS_PER_SEC);
#else
#define VAR(a,b) __typeof(b) a=(b)
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef unsigned long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\n")
#define I_de(x,n)for(int i=0;i<n;i++)printf("%d ",x[i]);Pln()
#define De(x)printf(#x"%d\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\
    printf("%d ",dp[htx][hty]);}Pln();}
#define M 30005
#define PII pair<int,int>
#define PB push_back
#define oo INT_MAX
#define Set_oo 0x3f
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-6
clock_t start_time=clock(), end_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
    int tmp=min(x,y);
    return min(tmp,z);
}
int max3(int x,int y,int z){
    int tmp=max(x,y);
    return max(tmp,z);
}
Int n;
Int mod=100000000;
Int bs(Int k){
    Int l=0,r=3037000499LL;
    while(r>=l){
        Int m=(r+l)/2;
//        debug("%lld\n",m);
        if(m*m>k)r=m-1;
        else l=m+1;
    }
    return r;
}
void solve(){
    if(n%2==0)n--;
    Int rt=bs(n);
    Int t=rt-1;
    Int ans=1LL;
    Int m[3]={t,t+1,2*t+1};
    for(int i=0,fg1=0,fg2=0;i<3;i++){
        if(m[i]%2==0&&!fg1){
            m[i]/=2;
            fg1=1;
        }
        if(m[i]%3==0&&!fg2){
            m[i]/=3;
            fg2=1;
        }
    }
//    Int ans2=(t*(t+1)*(2*t+1))/6;
//    ans2%=mod;
//    debug("%llu %llu %llu\n",ans,ans2,t);
    ans=(ans*m[0])%mod;
    ans=(ans*m[1])%mod;
    ans=(ans*m[2])%mod;
    Int nowdigit=t*(t+1)/2;
    Int x;
    if(t&1LL){
        x=t/2+1;
        nowdigit+=t/2+1;
    }else{
        x=t/2;
        nowdigit+=t/2;
    }
    x=(x*x);
    ans=(ans+x%mod)%mod;
//    debug("%lld\n",ans);
    Int limit=nowdigit*2-1;
    Int num=(n-limit)/2;
    ans=(ans+((num%mod)*rt)%mod)%mod;
//    debug("%lld %lld\n",num,rt);
//    debug("%lld %lld %lld\n",limit,num,x);
    printf("%lld\n",ans);
}
int main() {
    ios_base::sync_with_stdio(0);
    while(~scanf("%llu",&n)&&n){
        solve();
    }

















}