16.7.21

//今日代码

2016-07-21 training

（A题虽然比赛时没有打出来orz 题意是有7个英雄打3个怪，要分成3队求满足题意的最优。解法是因为数据很小直接for7遍，枚举每个人在哪个队。7个for的写法有点难看，另一种就是1个for从1到3^7，表示成3进制就能得出每个人在哪个队伍。）

B. Falling Anvils

（保证方程有实根则有p - 4q >= 0，答案即给定坐标范围内的几何概率）

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<map>
#define LL long long
#define INF 0x1f1f1f1f
using namespace std;

int main()
{
freopen("xx.in","r",stdin);
freopen("xx.out","w",stdout);

int t;
cin >> t;
while(t--)
{
double a,b;
scanf("%lf%lf",&a, &b);

double y = a/4;
double ans = 0;
double sum = 2*b*a;
double ans1 = (a*y)/2+a*b;
double ans2 = ((a-4*b)+a)*b/2+a*b;

if(b == 0) cout << 1 << endl;
else if(a == 0) cout << 0.5 << endl;
else
{
if(y<=b)
ans = ans1/sum;
else
ans = ans2/sum;

cout << ans << endl;
}
}

return 0;
}

D. Panoramix’s Prediction

（求b是否是a的下一个相邻素数）

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<map>
#define LL long long
#define INF 0x1f1f1f1f
using namespace std;

bool check(int num)//ispri
{
for(int i = 2; i<= sqrt(num) ;i++)
if(num%i == 0) return false;
return true;
}

int main()
{
freopen("xx.in","r",stdin);
freopen("xx.out","w",stdout);

int a,b;
cin >> a >> b;

if(check(b) == false)
cout << "NO" << endl;
else
{
bool flag = false;
for(int i = a+1; i < b; i++)
{
if(check(i) == true)
{
cout << "NO" << endl;
flag = true;
break;
}
}
if(!flag) cout << "YES" <<endl;
}

return 0;
}

E. Depression

（从00:00顺时针拨到给定时间，分针和时针需要转过的角度。注意4:30这样时针指在4和5之间。）

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<map>
#define LL long long
#define INF 0x1f1f1f1f
using namespace std;

int main()
{
freopen("xx.in","r",stdin);
freopen("xx.out","w",stdout);

double h,m;
scanf("%lf:%lf",&h,&m);

h+=m/60;
if(h>=12) h-=12;

cout << h/12*360 <<" "<<m/60*360<< endl;

return 0;
}

F. Magical Array

（连续区间中的一串数字的最大最小值相等则被认为是有魔力的，求有魔力的子串个数。即求区间连续且每个数字都相等的子串个数。可知相邻相等个数为n时答案即为n+（n-1）+…+1）

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<map>
#define LL long long
#define INF 0x1f1f1f1f
using namespace std;

LL arr[100010];
map<int, int> mm;

int main()
{
freopen("xx.in","r",stdin);
freopen("xx.out","w",stdout);

int n;
cin >> n;
for(int i = 0; i < n; i++)
scanf("%I64d",&arr[i]);

LL ans = 0;
for(int i = 0; i < n;)
{
LL cnt = 1;
int j = i+1;
while(arr[i] == arr[j] && j < n)
{
cnt++;
j++;
}
ans+=(cnt+1)*cnt/2;
i = j;
}

cout << ans << endl;

return 0;
}

I. Toy Army

（因为都是偶数，显然要保证被击中的人数最多，则双方各用一半的人数击打对方一半人，最终留下1/4的人。//请允许给出题人一个微笑脸。）

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<map>
#define LL long long
#define INF 0x1f1f1f1f
using namespace std;

int main()
{
freopen("xx.in","r",stdin);
freopen("xx.out","w",stdout);

LL n;
cin >> n;

cout << 2*n/4*3 << endl;

return 0;
}