| Educational Codeforces Round 26 |
|---|
You are given a text of single-space separated words, consisting of small and capital Latin letters.
Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text.
Calculate the volume of the given text.
The first line contains one integer number n (1 ≤ n ≤ 200) — length of the text.
The second line contains text of single-space separated words s1, s2, ..., si, consisting only of small and capital Latin letters.
Print one integer number — volume of text.
7
NonZERO
5
24
this is zero answer text
0
24
Harbour Space University
1
In the first example there is only one word, there are 5 capital letters in it.
In the second example all of the words contain 0 capital letters.
求一个单词内最多的大写字母个数，遇到空格处理下，最后也处理下
#include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; getchar(); string s; getline(cin,s); int sum=0,f=0; for(int i=0;s[i];i++){ if(s[i]==' '){ if(f>sum)sum=f; f=0; }else if(s[i]<'a')f++; } if(f>sum)sum=f; cout<<sum<<endl; return 0; }
The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each color should be used in exactly one stripe.
You are given a field n × m, consisting of characters 'R', 'G' and 'B'. Output "YES" (without quotes) if this field corresponds to correct flag of Berland. Otherwise, print "NO" (without quotes).
The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field.
Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes).
6 5
RRRRR
RRRRR
BBBBB
BBBBB
GGGGG
GGGGG
YES
4 3
BRG
BRG
BRG
BRG
YES
6 7
RRRGGGG
RRRGGGG
RRRGGGG
RRRBBBB
RRRBBBB
RRRBBBB
NO
4 4
RRRR
RRRR
BBBB
GGGG
NO
The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1.
怎样才是波兰的国旗，就是有RGB对吧，那我MAP一下枚举就好了啊
#include<bits/stdc++.h> using namespace std; char s[105][105]; int main() { int n,m; cin>>n>>m; for(int i=0;i<n;i++){ cin>>s[i]; } map<char,int>M; M['R']=0,M['B']=0,M['G']=0; if(n%3==0){ for(int i=0;i<n;i+=n/3){ char c=s[i][0]; for(int j=i;j<i+n/3;j++) for(int k=0;k<m;k++) if(c!=s[j][k]) M[c]=-1; if(M[c]==0){M[c]=1;} } } if(M['R']==1&&M['G']==1&&M['B']==1) return 0*puts("YES"); M['R']=0,M['B']=0,M['G']=0; if(m%3==0){ for(int i=0;i<m;i+=m/3){ char c=s[0][i]; for(int j=i;j<i+m/3;j++) for(int k=0;k<n;k++){ if(c!=s[k][j]) M[c]=-1;} if(M[c]==0)M[c]=1; } } if(M['R']==1&&M['G']==1&&M['B']==1) return 0*puts("YES"); puts("NO"); return 0; }
One very important person has a piece of paper in the form of a rectangle a × b.
Also, he has n seals. Each seal leaves an impression on the paper in the form of a rectangle of the size xi × yi. Each impression must be parallel to the sides of the piece of paper (but seal can be rotated by 90 degrees).
A very important person wants to choose two different seals and put them two impressions. Each of the selected seals puts exactly one impression. Impressions should not overlap (but they can touch sides), and the total area occupied by them should be the largest possible. What is the largest area that can be occupied by two seals?
The first line contains three integer numbers n, a and b (1 ≤ n, a, b ≤ 100).
Each of the next n lines contain two numbers xi, yi (1 ≤ xi, yi ≤ 100).
Print the largest total area that can be occupied by two seals. If you can not select two seals, print 0.
2 2 2
1 2
2 1
4
4 10 9
2 3
1 1
5 10
9 11
56
3 10 10
6 6
7 7
20 5
0
In the first example you can rotate the second seal by 90 degrees. Then put impression of it right under the impression of the first seal. This will occupy all the piece of paper.
In the second example you can't choose the last seal because it doesn't fit. By choosing the first and the third seals you occupy the largest area.
In the third example there is no such pair of seals that they both can fit on a piece of paper.
C也直接枚举就完事了，枚举左上，再判断另一个合适么？
等换了显示器我就可以换成这种大括号换行了
#include <bits/stdc++.h> using namespace std; int N,X,Y,a[105],b[105],f=0; void la(int x0,int y0,int x1,int y1) { if((x0+x1<=X)&&(y0<=Y)&&(y1<=Y)) { f=max(f,x0*y0+x1*y1); } if((y0+y1<=Y)&&(x0<=X)&&(x1<=X)) { f=max(f,x0*y0+x1*y1); } } int main() { cin>>N>>X>>Y; for(int i=0; i<N; i++) { cin>>a[i]>>b[i]; } for(int i=0; i<N; i++) { for(int j=0; j<i; j++) { la(a[i],b[i],a[j],b[j]); la(a[i],b[i],b[j],a[j]); la(b[i],a[i],a[j],b[j]); la(b[i],a[i],b[j],a[j]); } } cout<<f<<endl; return 0; }
Vasya is studying number theory. He has denoted a function f(a, b) such that:
- f(a, 0) = 0;
- f(a, b) = 1 + f(a, b - gcd(a, b)), where gcd(a, b) is the greatest common divisor of a and b.
Vasya has two numbers x and y, and he wants to calculate f(x, y). He tried to do it by himself, but found out that calculating this function the way he wants to do that might take very long time. So he decided to ask you to implement a program that will calculate this function swiftly.
The first line contains two integer numbers x and y (1 ≤ x, y ≤ 1012).
Print f(x, y).
3 5
3
6 3
1
数论的题目，想起来是没有是什么，因为好像和公因数有关，但又不知道有哪些关系还是看大佬代码比较靠谱
#include <bits/stdc++.h> using namespace std; typedef long long LL; const LL INF=1e18; vector <LL> s; LL gcd(LL a,LL b) { return b==0?a:gcd(b,a%b); } void la(LL x) { LL rx=sqrt(x+0.5); if(!(x&1)) { s.push_back(2); while(!(x&1)) x/=2; } for(int i=3; i<=rx; i+=2) { if(x%i==0) { s.push_back(i); while(x%i==0) x/=i; } } if(x>1) s.push_back(x); } int main() { LL x,y; scanf("%lld%lld",&x,&y); la(x); LL ans=0,m; while(y) { LL g=gcd(x,y); x/=g; y/=g; m=y; for(auto p:s) if(x%p==0) m=min(m,y%p); ans+=m; y-=m; } printf("%lld\n",ans); return 0; }
Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.
The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).
Print maximal roundness of product of the chosen subset of length k.
3 2
50 4 20
3
5 3
15 16 3 25 9
3
3 3
9 77 13
0
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
k个数的后缀0最多，10的因子是2和5，然后每次取2和5的最小值就好的，dp
#include<bits/stdc++.h> using namespace std; #define ll __int64 ll a[250]; int n2[250]; int n5[250]; int dp[2][205][12850]; int la2(ll num) { int sum=0; while(num%2==0)num/=2,sum++; return sum; } int la5(ll num) { int sum=0; while(num%5==0)num/=5,sum++; return sum; } int main() { int n,m; while(~scanf("%d%d",&n,&m)) { memset(dp,-1,sizeof(dp)); for(int i=1; i<=n; i++)scanf("%I64d",&a[i]); for(int i=1; i<=n; i++) { n2[i]=la2(a[i]); n5[i]=la5(a[i]); } int ma=0; dp[0][0][0]=0; int f1=0,f2=1; for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++) { for(int k=12805; k>=0; k--) { dp[f2][j][k]=max(dp[f1][j][k],dp[f1][j][k]); if(k>=n2[i]&&dp[f1][j-1][k-n2[i]]!=-1) dp[f2][j][k]=max(dp[f2][j][k],dp[f1][j-1][k-n2[i]]+n5[i]); ma=max(ma,min(k,dp[f2][j][k])); } } dp[f2][0][0]=dp[f1][0][0]; swap(f1,f2); } printf("%d\n",ma); } }