Codeforces Round #415 (Div. 2) 翻车啦】的更多相关文章

A. Straight «A» time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year.…
A. Straight «A» time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year.…
A:签到 #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; int read() { ,f=;char c=getchar(); ;c=getchar();} )+(x<<)+(c^),c=getchar(); ret…
A. Straight «A» 题面 Noora is a student of one famous high school. It's her final year in school - she is going to study in university next year. However, she has to get an «A» graduation certificate in order to apply to a prestigious one. In school, w…
A:签到 #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; int read() { ,f=;char c=getchar(); ;c=getchar();} )+(x<<)+(c^),c=getchar(); ret…
CodeForces - 810A A. Straight «A» time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Noora is a student of one famous high school. It's her final year in school - she is going to study in univ…
A:考虑每对最大值最小值的贡献即可. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 300010 #define P 1000000007 char getc(){…
A:签到. B:仅当只有一种字符时无法构成非回文串. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 1010 ')) c=getchar();return c;}…
反正又是一个半小时没做出来... 先排序,然后求和,第i个和第j个,f(a)=a[j]-a[i]=a[i]*(2^(j-i-1))因为从j到i之间有j-i-1个数(存在或者不存在有两种情况) 又有a[i+k]-a[i]=a[n]+a[n-1]+...+a[n-k]-a[k+1]-...-a[1] #include<map> #include<set> #include<cmath> #include<queue> #include<stack>…
A:签到. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long ')) c=getchar();return c;} ?n:gcd(m,n%m);} int read() { ,f=;cha…