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| Time Limit: 3000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
I was trying to solve problem ‘1234 - Harmonic Number‘, I wrote the following code
long long H( int n ) {
long long res = 0;
for( int i = 1; i <= n; i++ )
res = res + n / i;
return res;
}
Yes, my error was that I was using the integer divisions only. However, you are given n, you have to find H(n) as in my code.
Input
Input starts with an integer T (≤ 1000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n < 231).
Output
For each case, print the case number and H(n) calculated by the code.
Sample Input
11
1
2
3
4
5
6
7
8
9
10
2147483647
Sample Output
Case 1: 1
Case 2: 3
Case 3: 5
Case 4: 8
Case 5: 10
Case 6: 14
Case 7: 16
Case 8: 20
Case 9: 23
Case 10: 27
Case 11: 46475828386
Source
/** 题意：求n/1 + n/2 + n/3 + ........n/n 做法：先求tt = sqrt(n); 比如 n = 10 sqrt(10) = 3,10/1 ~ 10/10 之间有 10/6 ~ 10/10 之间都是1 个数为(10/1 - 10/2)个 之间为2的个数为 (10/4 ~ 10/5) 有(10/2 ~ 10/3) 个 为3的个数有 10 /3 有(10/3) 个 然后加判断 如果 n/(int)sqrt(n) == (int)sqrt(n) 则sum -= n/(int)sqrt(n) **/ #include<iostream> #include<cmath> #include<algorithm> #include<string.h> #include<stdio.h> using namespace std; int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif // ONLINE_JUDGE int T; int Case = 1; scanf("%d",&T); while(T--) { long long n; scanf("%lld",&n); long long res = 0; int tt = (int)sqrt(n); for(int i=1;i<=tt;i++) { res += (n/i); } for(int j=1;j<=tt;j++) { res += (n/j - n/(j+1)) * j; } if( n/(int)sqrt(n) == (int)sqrt(n)) res -= (n/(int)sqrt(n)); printf("Case %d: %lld\n",Case++,res); } return 0; }
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原文地址：http://www.cnblogs.com/chenyang920/p/4445544.html
