LightOJ 1234 Harmonic Number (打表)
2016-07-03 11:26
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Harmonic Number
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status Practice LightOJ 1234
Description
In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:
In this problem, you are given n, you have to find Hn.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 108).
Output
For each case, print the case number and the nth harmonic number. Errors less than 10-8 will be ignored.
Sample Input
12
1
2
3
4
5
6
7
8
9
90000000
99999999
100000000
Sample Output
Case 1: 1
Case 2: 1.5
Case 3: 1.8333333333
Case 4: 2.0833333333
Case 5: 2.2833333333
Case 6: 2.450
Case 7: 2.5928571429
Case 8: 2.7178571429
Case 9: 2.8289682540
Case 10: 18.8925358988
Case 11: 18.9978964039
Case 12: 18.9978964139
题意如题。
题解:考察超内存问题,要分组存储,否则会超内存。
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status Practice LightOJ 1234
Description
In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:
In this problem, you are given n, you have to find Hn.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 108).
Output
For each case, print the case number and the nth harmonic number. Errors less than 10-8 will be ignored.
Sample Input
12
1
2
3
4
5
6
7
8
9
90000000
99999999
100000000
Sample Output
Case 1: 1
Case 2: 1.5
Case 3: 1.8333333333
Case 4: 2.0833333333
Case 5: 2.2833333333
Case 6: 2.450
Case 7: 2.5928571429
Case 8: 2.7178571429
Case 9: 2.8289682540
Case 10: 18.8925358988
Case 11: 18.9978964039
Case 12: 18.9978964139
题意如题。
题解:考察超内存问题,要分组存储,否则会超内存。
#include <iostream> #include <cmath> #include <cstdio> using namespace std; const int maxn=1e8+5; double a[maxn/1000+5]; //注意+5 void get() { double sum=1.0; a[0]=0; a[1]=1.0; for(int i=2;i<=maxn;i++) { sum+=1.0/double(i); if(i%1000==0) a[i/1000]=sum; } } int main() { get(); int t,cas=1; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); int b=n/1000; double ans=a[b]; for(int i=b*1000+1;i<=n;i++) //注意是b*1000 ans+=1.0/double(i); printf("Case %d: %.10lf\n",cas++,ans); } return 0; }
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