ZOJ 3175 Number of Containers 分块

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3216

乱搞的...watashi是分块做的...但我并不知道什么是分块...大概就是把结果相同的数据合并计算

打表跑了一下...发现重复出现的数字很多...于是直接找出会发生重复的数乘起来就行了...

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define EPS             1e-8
#define DINF            1e15
#define MAXN            1000050
#define MOD             1000000007
#define INF             0x7fffffff
#define LINF            1LL<<60
#define PI              3.14159265358979323846
#define lson            l,m,rt<<1
#define rson            m+1,r,rt<<1|1
#define BUG             cout<<" BUG! "<<endl;
#define LINE            cout<<" ------------------ "<<endl;
#define FIN             freopen("in.txt","r",stdin);
#define FOUT            freopen("out.txt","w",stdout);
#define mem(a,b)        memset(a,b,sizeof(a))
#define FOR(i,a,b)      for(int i = a ; i < b ; i++)
#define read(a)         scanf("%d",&a)
#define read2(a,b)      scanf("%d%d",&a,&b)
#define read3(a,b,c)    scanf("%d%d%d",&a,&b,&c)
#define write(a)        printf("%d\n",a)
#define write2(a,b)     printf("%d %d\n",a,b)
#define write3(a,b,c)   printf("%d %d %d\n",a,b,c)
#pragma comment         (linker,"/STACK:102400000,102400000")
template<class T> inline T L(T a)       {return (a << 1);}
template<class T> inline T R(T a)       {return (a << 1 | 1);}
template<class T> inline T lowbit(T a)  {return (a & -a);}
template<class T> inline T Mid(T a,T b) {return ((a + b) >> 1);}
template<class T> inline T gcd(T a,T b) {return b ? gcd(b,a%b) : a;}
template<class T> inline T lcm(T a,T b) {return a / gcd(a,b) * b;}
template<class T> inline T Min(T a,T b) {return a < b ? a : b;}
template<class T> inline T Max(T a,T b) {return a > b ? a : b;}
template<class T> inline T Min(T a,T b,T c)     {return min(min(a,b),c);}
template<class T> inline T Max(T a,T b,T c)     {return max(max(a,b),c);}
template<class T> inline T Min(T a,T b,T c,T d) {return min(min(a,b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d) {return max(max(a,b),max(c,d));}
template<class T> inline T exGCD(T a, T b, T &x, T &y){
    if(!b) return x = 1,y = 0,a;
    T res = exGCD(b,a%b,x,y),tmp = x;
    x = y,y = tmp - (a / b) * y;
    return res;
}
template<class T> inline T reverse_bits(T x){
    x = (x >> 1 & 0x55555555) | ((x << 1) & 0xaaaaaaaa); x = ((x >> 2) & 0x33333333) | ((x << 2) & 0xcccccccc);
    x = (x >> 4 & 0x0f0f0f0f) | ((x << 4) & 0xf0f0f0f0); x = ((x >> 8) & 0x00ff00ff) | ((x << 8) & 0xff00ff00);
    x = (x >>16 & 0x0000ffff) | ((x <<16) & 0xffff0000); return x;
}
typedef long long LL;    typedef unsigned long long ULL;
//typedef __int64 LL;      typedef unsigned __int64 ULL;

/*********************   By  F   *********************/
int main(){
    //FIN;
    //FOUT;
    int T;
    while(cin>>T){
        while(T--){
            LL n,res = 0;
            cin>>n;
            if(n == 1){
                cout<<"0"<<endl;
                continue;
            }
            LL tmp = res = n;
            for(LL i = 2 ; i <= n ; i++){
                if(tmp - n/i <= 1) {
                    res -= tmp;
                    break;
                }
                res += n/i;
                tmp = n/i;
            }
            LL t = n;
            for(LL i = 1 ; i <= tmp ; i++){
                res += (t - n/(i+1)) * i ;
                t = n/(i+1);
            }
            res -= n;
            cout<<res<<endl;
        }
    }
    return 0;
}

 

posted @ 2013-10-06 20:30  Felix_F  阅读(229)  评论(0编辑  收藏  举报