lightoj Harmonic Number (II) 1245 (数论)
2015-11-07 21:35
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Harmonic Number (II)
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
//很好的题解:
先求出前sqrt(n)项和:即n/1+n/2+...+n/sqrt(n)
再求出后面所以项之和.后面每一项的值小于sqrt(n),计算值为1到sqrt(n)的项的个数,乘以其项值即可快速得到答案
例如:10/1+10/2+10/3+...+10/10
sqrt(10) = 3
先求出其前三项的和为10/1+10/2+10/3
在求出值为1的项的个数为(10/1-10/2)个,分别是(10/10,10/9,10/8,10/7,10/6),值为2个项的个数(10/2-10/3)分别是(10/5,10/4),在求出值为3即sqrt(10)的项的个数.
显然,值为sqrt(10)的项计算了2次,减去一次即可得到答案。当n/(int)sqrt(n) == (int)sqrt(n)时,值为sqrt(n)的值会被计算2次。
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
//很好的题解:
先求出前sqrt(n)项和:即n/1+n/2+...+n/sqrt(n)
再求出后面所以项之和.后面每一项的值小于sqrt(n),计算值为1到sqrt(n)的项的个数,乘以其项值即可快速得到答案
例如:10/1+10/2+10/3+...+10/10
sqrt(10) = 3
先求出其前三项的和为10/1+10/2+10/3
在求出值为1的项的个数为(10/1-10/2)个,分别是(10/10,10/9,10/8,10/7,10/6),值为2个项的个数(10/2-10/3)分别是(10/5,10/4),在求出值为3即sqrt(10)的项的个数.
显然,值为sqrt(10)的项计算了2次,减去一次即可得到答案。当n/(int)sqrt(n) == (int)sqrt(n)时,值为sqrt(n)的值会被计算2次。
#include<stdio.h> #include<string.h> #include<math.h> int main() { int t,n,i,j; int T=1; scanf("%d",&t); while(t--) { scanf("%d",&n); long long ans=0; for(i=1;i<=(int)sqrt(n);i++) { ans+=n/i; if(n/i>n/(i+1)) ans+=(long long)((n/i-n/(i+1))*i); } i--; if(n/i==i) ans-=i; printf("Case %d: %lld\n",T++,ans); } return 0; }
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