标签：
What is the value this simple C++ function will return?
1 long long H(int n) 2 3 { 4 long long res = 0; 5 for( int i = 1; i <= n; i=i+1 ) 6 7 { 8 res = (res + n/i); 9 } 10 return res; 11 }
Input
The first line of input is an integer T (T ≤ 1000) that indicates the number of test cases. Each of the next T line will contain a single signed 32 bit integer n.
Output
For each test case, output will be a single line containing H(n).
Sample Input
2
5
10
Sample Output
10
27
题意：求上述函数的返回值；
分析：简单讲就是求出一个tmp=[n/i]后计算一下这个数会出现多少次，方法就是n/tmp,求得的数是满足n/i==tmp的最大i值，然后继续更新i值即可。算法复杂度会由O(n)降为O(sqrt(n));
代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cmath> 6 #include<algorithm> 7 using namespace std; 8 typedef long long LL; 9 LL H(LL n) 10 { 11 LL res = 0; 12 LL pre = n, next_pos = n, tmp; 13 for(LL i = 1; i <= n; i++) 14 { 15 res += n/i; 16 tmp = n/i; 17 if(tmp != pre) 18 { 19 next_pos = n/tmp; 20 res += (next_pos-i)*tmp; 21 pre = tmp; 22 i = next_pos; 23 } 24 } 25 return res; 26 } 27 int main() 28 { 29 int T; scanf("%d", &T); 30 while(T--) 31 { 32 LL n; scanf("%lld", &n); 33 printf("%lld\n", H(n)); 34 } 35 return 0; 36 }
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原文地址：http://www.cnblogs.com/LLGemini/p/4327016.html
