2016 ACMICPC亚洲区域青岛在线B题的解题思路是什么？
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本文共计529个文字，预计阅读时间需要3分钟。
python简化后的代码
def sum_of_reciprocal_squares(n): return sum(1 / i**2 for i in range(1, n + 1))
读取输入并处理多个测试案例import sysinput=sys.stdin.readdata=input().split()
for line in data: if line.strip(): n=int(line.strip()) result=sum_of_reciprocal_squares(n) print(result)
Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.
Input
There are multiple cases.
For each test case, there is a single line, containing a single positive integer n.
The input file is at most 1M.
Output
The required sum, rounded to the fifth digits after the decimal point.
Sample Input
1
2
4
8
15
Sample Output
1.00000
1.25000
1.42361
1.52742
1.58044
Σ(1/n^2) 的极限为 pi^2/6。。。。。当然是趋近很大的时候，所以简单设置一下上界即可=。= （送分题）
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 8000005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-7
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
a = a % mod;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
double sum[maxn];
int main()
{
//ios::sync_with_stdio(false);
string s;
for (int i = 1; i <= 1000000; i++) {
sum[i] = sum[i - 1] + 1.0 / i / i;
}
while (cin>>s) {
int len = s.length();
//printf("%.5f\n", 1.0*pi*pi / 6);
if (len >= 7)
printf("%.5f\n", 1.0*pi*pi / 6);
else {
int n = 0;
for (int i = 0; i < len; i++) {
n = n * 10 + s[i] - '0';
}
printf("%.5f\n", 1.00*sum[n]);
}
}
}
本文共计529个文字，预计阅读时间需要3分钟。
python简化后的代码
def sum_of_reciprocal_squares(n): return sum(1 / i**2 for i in range(1, n + 1))
读取输入并处理多个测试案例import sysinput=sys.stdin.readdata=input().split()
for line in data: if line.strip(): n=int(line.strip()) result=sum_of_reciprocal_squares(n) print(result)
Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.
Input
There are multiple cases.
For each test case, there is a single line, containing a single positive integer n.
The input file is at most 1M.
Output
The required sum, rounded to the fifth digits after the decimal point.
Sample Input
1
2
4
8
15
Sample Output
1.00000
1.25000
1.42361
1.52742
1.58044
Σ(1/n^2) 的极限为 pi^2/6。。。。。当然是趋近很大的时候，所以简单设置一下上界即可=。= （送分题）
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 8000005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-7
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
a = a % mod;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
double sum[maxn];
int main()
{
//ios::sync_with_stdio(false);
string s;
for (int i = 1; i <= 1000000; i++) {
sum[i] = sum[i - 1] + 1.0 / i / i;
}
while (cin>>s) {
int len = s.length();
//printf("%.5f\n", 1.0*pi*pi / 6);
if (len >= 7)
printf("%.5f\n", 1.0*pi*pi / 6);
else {
int n = 0;
for (int i = 0; i < len; i++) {
n = n * 10 + s[i] - '0';
}
printf("%.5f\n", 1.00*sum[n]);
}
}
}