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Educational Codeforces Round 60 (Rated for Div. 2)E(思维,哈希,字符串,交互)(哈希算法思想)

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本文将带您了解关于Educational Codeforces Round 60 (Rated for Div. 2)E的新内容,同时我们还将为您解释思维,哈希,字符串,交互的相关知识,另外,我们还将为您提供关于Educational Codeforces Round 33 (Rated for Div. 2)、Educational Codeforces Round 36 (Rated for Div. 2) A-C、Educational Codeforces Round 37 (Rated for Div. 2)、Educational Codeforces Round 41 (Rated for Div. 2)的实用信息。

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Educational Codeforces Round 60 (Rated for Div. 2)E(思维,哈希,字符串,交互)(哈希算法思想)

Educational Codeforces Round 60 (Rated for Div. 2)E(思维,哈希,字符串,交互)(哈希算法思想)

#include <bits/stdc++.h>using namespace std;int main(){ string t; cin>>t; int n=t.size(); string s1(n,‘a‘),s2(n,s3(n,‘a‘); for(int i=0;i<n;i++){  s1[i]=char(‘a‘+(i%26));//从a到z循环  s2[i]=char(‘a‘+((i/26)%26));//第i位为(i/26)%26+‘a‘,保证了26*26内不会和s1相同  s3[i]=char(‘a‘+((i/26/26)%26));//保证了26^3内不会和s1相同,类似哈希编码 } cout<<"? "<<s1<<endl; string t1; cin>>t1; cout<<"? "<<s2<<endl; string t2; cin>>t2; cout<<"? "<<s3<<endl; string t3; cin>>t3; vector<int>p(n); for(int i=0;i<n;i++)  p[i]=(t1[i]-‘a‘)+(t2[i]-‘a‘)*26+(t3[i]-‘a‘)*26*26;//将变化后的值根据编码还原 string s(n,‘a‘); for(int i=0;i<n;i++)  s[p[i]]=t[i];//26^3>1000,所以可行 cout<<"! "<<s<<endl; return 0;}

Educational Codeforces Round 33 (Rated for Div. 2)

Educational Codeforces Round 33 (Rated for Div. 2)

总结

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Educational Codeforces Round 36 (Rated for Div. 2) A-C

Educational Codeforces Round 36 (Rated for Div. 2) A-C

总结

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Educational Codeforces Round 37 (Rated for Div. 2)

Educational Codeforces Round 37 (Rated for Div. 2)

总结

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Educational Codeforces Round 41 (Rated for Div. 2)

Educational Codeforces Round 41 (Rated for Div. 2)

总结

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