Educational Codeforces Round 64 (Rated for Div. 2)
2019-05-03 21:52
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A. Inscribed Figures
题意:就是1 2 3分别代表 圆 等腰三角形 正方形,然后往每一个图形里面套下一个图形,求最后的相交点总共有多少个(题意还是有点难懂的,(我不会说是他题目一开始改来改去的我有点懵逼),其实是我菜吧。。。)
分析:
# include <bits/stdc++.h> using namespace std; typedef long long LL; int a[110]; LL sum=0; int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%d",&a[i]); } for(int i=1;i<n;i++){ if(a[i-1]==1){ if(a[i]==2){ if((i-2)>=0&&a[i-2]==3){ sum+=2; }else{ sum+=3; } }else if(a[i]==3){ sum+=4; } }else if(a[i-1]==2){ if(a[i]==1){ sum+=3; }else if(a[i]==3){ printf("Infinite"); return 0; } }else if(a[i-1]==3){ if(a[i]==1){ sum+=4; }else if(a[i]==2){ printf("Infinite"); return 0; } } } printf("Finite\n%lld",sum); return 0; }
B - Ugly Pairs
可以按照奇书偶数来把字符排序,然后最后再判断这个顺序是否会造成NO的情况
有一个可能会造成思维定势的地方就是,可以先拍好序列再判断是否会造成NO的情况,而不是先判断是否会造成NO的情况再排序(可能大佬们没有这个困惑QWQ)
# include <bits/stdc++.h> using namespace std; int main() { int t; string s; string a,b; scanf("%d",&t); while(t--){ a="",b=""; cin>>s; sort(s.begin(),s.end()); for(int i=0;i<s.length();i++){ if(s[i]%2) a+=s[i]; else b+=s[i]; } if(abs(a[a.length()-1]-b[0])!=1){ cout<<a<<b<<endl; }else if(abs(b[b.length()-1]-a[0])!=1){ cout<<b<<a<<endl; }else{ cout<<"No answer"<<endl; } } return 0; }
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