https://www.cnblogs.com/31415926535x/p/10397007.html
codeforces 537 div2
A
题意就是给你两个字符串，然后如果s,t的对应位上的字母要么都是元音，要么都是辅音，，就输出Yes反之输出No，，长度不等肯定输出的是No，，，
#include
//#include
//#include
//#include
//#include
#define aaa cout< '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
bool check(char a, char b)
{
if(a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u')
if(b == 'a' || b == 'e' || b == 'i' || b == 'o' || b == 'u')
return true;
else
return false;
else if(b != 'a' && b != 'e' && b != 'i' && b != 'o' && b != 'u')return true;
else return false;
}
int main()
{
// freopen("233.in" , "r" , stdin);
// freopen("233.out" , "w" , stdout);
ios_base::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
string s, t;cin >> s >> t;
if(s.length() != t.length())cout << "No" << endl;
else
{
int len = s.length();
for(int i = 0; i < len; ++i)
{
if(!check(s[i], t[i]))
{
cout << "No" << endl;
return 0;
}
}
cout << "Yes" << endl;
}
return 0;
} B
题意是给你n个数，有两种操作，一个是删除任意的一个数，另一个是将任意的一个数加一，，对于 每个数的操作 最多有k种，，总的操作数是m，，，然后问你m个操作后最大的平均值是多少，，
首先为了尽可能的增加平均数，要删除一些小的数，，暴力遍历可能删除的数的个数，，显然最多删除的个数是n-1或者是m，，所以遍历的边界是 min(m, n - 1)，，
然后依次删去最小的数（预先排序一下），，删掉这个数后，算一下此时剩下数的平均值，，，然后和上一次的结果比较一下，取最大就行
#include
//#include
//#include
//#include
//#include
#define aaa cout<<233< pii;
const int inf = 0x3f3f3f3f;//1061109567
const ll linf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = 3.14159265358979;
const int maxn = 1e5 + 5;
const int maxm = 2e5 + 5;
const ll mod = 1e9 + 7;
inline int read() //快读
{
int ans=0;
char ch=getchar();
while(!isdigit(ch))
ch=getchar();
while(isdigit(ch))
ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
return ans;
}
ll a[maxn];
int main()
{
// freopen("233.in" , "r" , stdin);
// freopen("233.out" , "w" , stdout);
// ios_base::sync_with_stdio(0);
// cin.tie(0);cout.tie(0);
ll n, k, m;
n = read(); k = read(); m = read();
for(int i = 1; i <= n; ++i)a[i] = read();
sort(a + 1, a + 1 + n);
ll sum = 0;
for(int i = 1; i <= n; ++i)sum += a[i];
long double ans = (long double)(sum + min(k * n, m)) / (long double)(n);
for(int i = 1; i <= min(m, n - 1); ++i)
{
sum -= a[i];
long double res = (long double)(sum + min(m - i, k * (n - i))) / (long double)(n - i);
ans = max(ans, res);
}
printf("%.20f", (double)ans);
return 0;
} C
题意是给你一个区间长度为 \(2^n\)长，，然后一个数组a[k]，a[i]表示第i个位置加一，，可能有a[i]是相等的，，然后有两种操作，一种是子区间全为零时操作的代价为A，，否则代价为 \(B*num_{l,r}*len_{l, r}\)，，，问你整个区间的最小操作代价，，
题解是递归+二分求解，，，
我一开始想到了递归来求，，但是自己写二分求区间[l, r]的 \(num_{l, r}\) 时总是写爆，，，最后看了题解才想起来还有stl里的 lower_bound 和 upper_bound 可以直接二分找到，，，QAQ
#include
//#include
//#include
//#include
//#include
#define aaa cout<<233< pii;
const int inf = 0x3f3f3f3f;//1061109567
const ll linf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = 3.14159265358979;
const int maxn = 1e5 + 5;
const int maxm = 2e5 + 5;
const ll mod = 1e9 + 7;
inline ll read() //快读
{
ll ans=0;
char ch=getchar();
while(!isdigit(ch))
ch=getchar();
while(isdigit(ch))
ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
return ans;
}
vector a;
ll n, k, A, B;
#define len (r-l+1)
#define mid ((l+r)>>1)
ll getnum(int l, int r)
{
l = lower_bound(a.begin(), a.end(), l) - a.begin();
r = upper_bound(a.begin(), a.end(), r) - a.begin();
return r - 1 - l + 1;
}
ll solve(int l, int r)
{
ll num = getnum(l, r);
if(!num)return A;
if(l == r)
{
if(num)
return B * num * 1;
else
return A;
}
ll a = solve(l, mid);
ll b = solve(mid + 1, r);
// cout << a << b << "---" << endl;
if(num)return min(a+b, (ll)(B * len * num));
else return min(a+b, A);
}
int main()
{
// freopen("233.in" , "r" , stdin);
// freopen("233.out" , "w" , stdout);
// ios_base::sync_with_stdio(0);
// cin.tie(0);cout.tie(0);
n = read(); k = read(); A = read(); B = read();
for(int i = 1; i <= k; ++i)
{
int t = read();
a.pb(t);
}
sort(a.begin(), a.end());
printf("%lld", solve(1, (1<