1245 - Harmonic Number (II)(规律题)
2015-11-07 22:28
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1245 - Harmonic Number (II)
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.
denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤
n < 231).
by the code.
题解,自己有思路,却写不出来,看了答案,倒是挺简单的;
先看两个例子
1.
n = 10 sqrt(10) = 3 10/sqrt(10) = 3
i 1 2 3 4 5 6 7 8 9 10
n/i 10 5 3 2 2 1 1 1 1 1
m = n/i
sum += m;
m = 1的个数10/1-10/2 = 5;
m = 2的个数10/2-10/3 = 2;
m = 3的个数10/3-10/4 = 1;
2.
n = 20 sqrt(20) = 4 20/sqrt(20) = 5
i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
n/i 20 10 6 5 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1
m = n/i
sum += m;
m = 1的个数20/1-20/2 = 10;
m = 2的个数20/2-20/3 = 4;
m = 3的个数20/3-20/4 = 1;
m = 4的个数20/4-20/5 = 1;
...
m = i的个数20/i - 20/(i + 1)(1<= i <= sqrt(n))
这样我们可以得出:sqrt(n)之前的数我们可以直接用for循环来求
sqrt(n)之后的sum += (n/i - n/(i + 1)) * i;
当sqrt(n) = n / sqrt(n)时(如第一个例子10,sum就多加了一个3),sum多加了一个sqrt(n),减去即可;
无语,这里输出%I64d竟然wa。。。
代码:
PDF (English) | Statistics | Forum |
Time Limit: 3 second(s) | Memory Limit: 32 MB |
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) calculatedby the code.
Sample Input | Output for Sample Input |
11 1 2 3 4 5 6 7 8 9 10 2147483647 | 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 |
先看两个例子
1.
n = 10 sqrt(10) = 3 10/sqrt(10) = 3
i 1 2 3 4 5 6 7 8 9 10
n/i 10 5 3 2 2 1 1 1 1 1
m = n/i
sum += m;
m = 1的个数10/1-10/2 = 5;
m = 2的个数10/2-10/3 = 2;
m = 3的个数10/3-10/4 = 1;
2.
n = 20 sqrt(20) = 4 20/sqrt(20) = 5
i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
n/i 20 10 6 5 4 3 2 2 2 2 1 1 1 1 1 1 1 1 1 1
m = n/i
sum += m;
m = 1的个数20/1-20/2 = 10;
m = 2的个数20/2-20/3 = 4;
m = 3的个数20/3-20/4 = 1;
m = 4的个数20/4-20/5 = 1;
...
m = i的个数20/i - 20/(i + 1)(1<= i <= sqrt(n))
这样我们可以得出:sqrt(n)之前的数我们可以直接用for循环来求
sqrt(n)之后的sum += (n/i - n/(i + 1)) * i;
当sqrt(n) = n / sqrt(n)时(如第一个例子10,sum就多加了一个3),sum多加了一个sqrt(n),减去即可;
无语,这里输出%I64d竟然wa。。。
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #define mem(x,y) memset(x,y,sizeof(x)) using namespace std; typedef long long LL; const int INF=0x3f3f3f3f; int main(){ int T; int n,cnt=0; LL res; scanf("%d",&T); while(T--){ res=0; scanf("%d",&n); int m=sqrt(n); for(int i=1;i<=m;i++)res+=n/i; for(int i=1;i<=m;i++) res+=(n/i-n/(i+1))*i; if(m==n/m)res-=m; printf("Case %d: %lld\n",++cnt,res); } return 0; }
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