Harmonic Number (II)(数论)
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
本题求Σn/i,i从1到n;
解法一 根号n之前的累加 m=Σn/i;n<sqrt(n),m*2-sqrr(n)*sqrt(n);我还不知道怎么推出来的
在这里插入代码片 #include<stdio.h> #include<math.h> int main() { int t,kk=1; scanf("%d",&t); while(t--) { long long n,sum=0; scanf("%lld",&n); int m=(int)sqrt(n); for(int i=1; i<=m; i++) sum=sum+n/i; sum*=2; sum-=m*m; printf("Case %d: %lld\n",kk++,sum); } return 0; }
解法二 估计也由一来的
因为从n/2 到 n 都是1 n/2 到 n/3 都是2 而 根号n之前的直接暴力累加
比如 13
根号13=3 ,13/1 + 13/2 +13/3 13/1-13/2(13,12,11,10,9,8,7) 13/2-13/3(6,5) 13/3-13/4 (4)
而 当根号n = n/根号n 时
比如 9 9/3=3 而 9/2-9/3 4 9/3-9/4(3),那么3 就算了两次 所以要减去。
在这里插入代码片 #include<math.h> #include<stdio.h> #include<string.h> int n,i,j,k,l,flag=1,t; int main() { scanf("%d",&t); while(t--) { scanf("%d",&n); long long k=sqrt(n);//求到2^16次方是不会超市的 long long m=0; for(i=1;i<k+1;i++) { m+=n/i;//前边的按正常计算 ,可计算到前1/k。 if(n/i>n/(i+1))//算了k个数后,开始找规律。 m+=(n/i-n/(i+1))*i;//从n到n/2这些值都为1,从n/2到n/3之间的数都为2...; } if(k==n/k) m=m-k; printf("Case %d: %lld\n",flag++,m); } return 0; }Starry_Sky_Dream 原创文章 50获赞 4访问量 2351 关注 私信
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