集训队训练赛--20170402

训练赛

地址

A - Circle in Square

[LightOJ 1022]

A circle is placed perfectly into a square. The term perfectly placed
means that each side of the square is touched by the circle, but the
circle doesn't have any overlapping part with the square. See the
picture below.

Now you are given the radius of the circle. You have to find the area
of the shaded region (blue part). Assume that pi = 2 * acos (0.0)
(acos means cos inverse).

Input

Input starts with an integer T (≤ 1000), denoting the number of test
cases.

Each case contains a floating point number r (0 < r ≤ 1000) denoting
the radius of the circle. And you can assume that r contains at most
four digits after the decimal point.

Output

For each case, print the case number and the shaded area rounded to
two places after the decimal point.

Sample Input

3
20
30.091
87.0921
Sample Output
Case 1: 343.36
Case 2: 777.26
Case 3: 6511.05

Hint

This problem doesn't have special judge. So, be careful about
precision problems. Better to add a small value to your result to
avoid precision problems. For example, add 10-9 to your result.

• 题意
  知道一个圆的半径，求以该圆直径为边长的正方形与该圆面积之差。
• 思路
  S=2r*2r-pi*r*r即可。
  注意此次取pi=2 * acos (0.0)，要使用acos函数。
• 代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
int main(){
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        double f,s,pi;
        pi=2.0*acos(0.0);
        scanf("%lf",&f);
        s=4*f*f-pi*f*f;
        printf("Case %d: %.2f\n",i,s);
    }
    return 0;
}

J - Intelligent Factorial Factorization

[LightOJ - 1035]

Given an integer N, you have to prime factorize N! (factorial N).

Input

Input starts with an integer T (≤ 125), denoting the number of test
cases.

Each case contains an integer N (2 ≤ N ≤ 100).

Output

For each case, print the case number and the factorization of the
factorial in the following format as given in samples.

Case x: N = p1 (power of p1) * p2 (power of p2) * ...

Here x is the case number, p1, p2 ... are primes in ascending order.

Sample Input

3
2
3
6

Sample Output

Case 1: 2 = 2 (1)
Case 2: 3 = 2 (1) * 3 (1)
Case 3: 6 = 2 (4) * 3 (2) * 5 (1)

• 题意
  给出整数N（2 ≤ N ≤ 100），将N的阶乘分解为质数相乘的形式。
• 思路
  从2到N依次分解成质数相乘的形式，而后合并。
• 代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
int ha[105];
int main(){
    int t,kase=0;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=2;i<=100;i++){
            ha[i]=0;
        }
        for(int i=2;i<=n;i++){
            int te=i;
            for(int j=2;j<=te;j++){
                if (te%j==0)
                {
                    te=te/j;
                    ha[j]++;
                    j--;
                }
            }
        }
        printf("Case %d: %d = ",++kase,n);
        int flag=0;
        for(int i=2;i<=n;i++){
            if (ha[i]==0)
            {
                continue;
            }
            if (flag)
            {
                printf(" * ");
            }
            flag=1;
            printf("%d (%d)",i,ha[i]);
        }
        printf("\n");
    }
    return 0;
}