LightOJ 1234 Harmonic Number(打表)

题目链接：https://vjudge.net/problem/LightOJ-1234

 

解题思路：

由于n的范围是1e8，暴力for循环累加，普通打表肯定会超。既然是较大后精度误差较小，那就提前每隔100提前记录一下和，每次输入n后再for循环单独累加不足100次即可！

 

AC代码：

#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <ctype.h>
//******************************************************
#define lrt (rt*2)
#define rrt  (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define exp 1e-8
//***************************************************
#define eps             1e-8
#define inf             0x3f3f3f3f
#define INF             2e18
#define LL              long long
#define ULL             unsigned long long
#define PI              acos(-1.0)
#define pb              push_back
#define mk              make_pair

#define all(a)          a.begin(),a.end()
#define rall(a)         a.rbegin(),a.rend()
#define SQR(a)          ((a)*(a))
#define Unique(a)       sort(all(a)),a.erase(unique(all(a)),a.end())
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(min(a,b),min(c,d))
#define max4(a,b,c,d)   max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define Iterator(a)     __typeof__(a.begin())
#define rIterator(a)    __typeof__(a.rbegin())
#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0)
#define CasePrint       pc('C'); pc('a'); pc('s'); pc('e'); pc(' '); write(qq++,false); pc(':'); pc(' ')
#define vi              vector <int>
#define vL              vector <LL>
#define For(I,A,B)      for(int I = (A); I < (B); ++I)
#define FOR(I,A,B)      for(int I = (A); I <= (B); ++I)
#define rFor(I,A,B)     for(int I = (A); I >= (B); --I)
#define Rep(I,N)        For(I,0,N)
#define REP(I,N)        FOR(I,1,N)
using namespace std;
const int maxn=1e6+5;


double a[maxn];

int main()
{
    int t,n,k=1;
    double sum=1.0;
    a[0]=0;
    a[1]=1;
    for(int i=2; i<=1e8; i++)
    {
        sum+=1.0/i;
        if(i%100==0) a[i/100]=sum;
    }

    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        int x=n/100;
        double sum=a[x];
        for(int i=x*100+1;i<=n;i++)
        {
            sum+=1.0/i;
        }
        printf("Case %d: %.10lf\n",k++,sum);
    }
    return 0;
}