链接：http://codeforces.com/contest/1265
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string ss, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition si≠si+1si≠si+1 should be satisfied for all 1≤i≤|s|−11≤i≤|s|−1, where |s||s| is the length of the string ss.
The first line contains positive integer tt (1≤t≤10001≤t≤1000) — the number of test cases. Next tt lines contain the descriptions of test cases.
Each line contains a non-empty string ss consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string ss has at least one character '?'. The sum of lengths of strings ss in all test cases does not exceed 105105.
For each test case given in the input print the answer in the following format:
- If it is impossible to create a beautiful string, print "-1" (without quotes);
- Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
3 a???cb a??bbc a?b?c
ababcb -1 acbac
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 44-th and 55-th characters will be always equal.
In the third test case, the only answer is "acbac".
题意很简单，给定一个只含a,b,c,?的字符串，如果字符串出现连续相同的字母(!='?')，就是输出-1,否则把？处填上字母，保证不会出现连续相同字母
自己写的实在是太麻烦，把几种情况都考虑进去了: 1：？？？？ 2：a????? 3:??????a 对比其他人的代码，感觉自己真是蠢，还是菜！
先上自己的渣渣代码：
#include<iostream> #include<cstring> using namespace std; const int maxn = 1e5+10; char s[maxn]; int main() { int t; cin>>t; while(t--) { cin>>s; int len = strlen(s); int ok = 0 ,ok2=0; for(int i = 0 ; i < len; i++) { if(s[i]==s[i+1]&&s[i]!='?') ok=1; if(s[i]!='?') ok2=1; } if(!ok2) { s[0]='a'; for(int i = 1 ; i < len ; i++) { if(s[i-1]=='c') s[i]='a'; else s[i]=s[i-1]+1; } cout<<s<<endl; continue ; } if(ok) { cout<<"-1"<<endl; continue; } for(int i = 0 ; i< len;i++) { if(s[i]=='?') { int l=i; int r; for(int j = i;j<len;j++) { if(s[j]!='?') break; r=j; // cout<<r<<endl; } //cout<<r<<" "<<len<<endl; if(i==0) { for(int j=r;j>=l;j--) { if(s[j+1]=='a') s[j]='c'; else s[j]=s[j+1]-1; } } else if(r==len-1) { for(int j=l;j<=r;j++) { if(s[j-1]=='c') s[j]='a'; else s[j]=s[j-1]+1; } } else { for(int j=l;j<r;j++) { if(s[j-1]=='c') s[j]='a'; else s[j]=s[j-1]+1; } if(s[r-1]=='a'&&s[r+1]=='a'||s[r-1]=='a'&&s[r+1]=='b'||s[r-1]=='b'&&s[r+1]=='a') s[r]='c'; else if(s[r-1]=='b'&&s[r+1]=='b'||s[r-1]=='b'&&s[r+1]=='c'||s[r-1]=='c'&&s[r+1]=='b'||s[r-1]=='c'&&s[r+1]=='c') s[r]='a'; else s[r]='b'; } } } cout<<s<<endl; } }
赛后参考别人然后自己写的代码：
#include<iostream> #include<cstring> using namespace std; const int maxn = 1e5+10; char s[maxn]; int main() { int t; cin>>t; while(t--) { cin>>s; int len=strlen(s); int ok = 0; for(int i =0 ; i< len ; i++) { if(s[i]==s[i+1]&&s[i]!='?') { ok=1;break; } if(s[i]=='?') { for(int j=0;j<3;j++) { if('a'+j!=s[i-1]&&'a'+j!=s[i+1]) //管他？在什么位置，直接中间不能于两边就上就行了。 { s[i]='a'+j;break; } } } } if(ok) cout<<"-1"<<endl; else cout<<s<<endl; } }
B-Beautiful Numbers
You are given a permutation p=[p1,p2,…,pn]p=[p1,p2,…,pn] of integers from 11 to nn. Let's call the number mm (1≤m≤n1≤m≤n) beautiful, if there exists two indices l,rl,r (1≤l≤r≤n1≤l≤r≤n), such that the numbers [pl,pl+1,…,pr][pl,pl+1,…,pr] is a permutation of numbers 1,2,…,m1,2,…,m.
For example, let p=[4,5,1,3,2,6]p=[4,5,1,3,2,6]. In this case, the numbers 1,3,5,61,3,5,6 are beautiful and 2,42,4 are not. It is because:
- if l=3l=3 and r=3r=3 we will have a permutation [1][1] for m=1m=1;
- if l=3l=3 and r=5r=5 we will have a permutation [1,3,2][1,3,2] for m=3m=3;
- if l=1l=1 and r=5r=5 we will have a permutation [4,5,1,3,2][4,5,1,3,2] for m=5m=5;
- if l=1l=1 and r=6r=6 we will have a permutation [4,5,1,3,2,6][4,5,1,3,2,6] for m=6m=6;
- it is impossible to take some ll and rr, such that [pl,pl+1,…,pr][pl,pl+1,…,pr] is a permutation of numbers 1,2,…,m1,2,…,m for m=2m=2 and for m=4m=4.
You are given a permutation p=[p1,p2,…,pn]p=[p1,p2,…,pn]. For all mm (1≤m≤n1≤m≤n) determine if it is a beautiful number or not.
The first line contains the only integer tt (1≤t≤10001≤t≤1000) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of the given permutation pp. The next line contains nnintegers p1,p2,…,pnp1,p2,…,pn (1≤pi≤n1≤pi≤n, all pipi are different) — the given permutation pp.
It is guaranteed, that the sum of nn from all test cases in the input doesn't exceed 2⋅1052⋅105.
Print tt lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length nn, there the ii-th character is equal to 11 if ii is a beautiful number and is equal to 00 if ii is not a beautiful number.
3 6 4 5 1 3 2 6 5 5 3 1 2 4 4 1 4 3 2
101011 11111 1001
The first test case is described in the problem statement.
In the second test case all numbers from 11 to 55 are beautiful:
- if l=3l=3 and r=3r=3 we will have a permutation [1][1] for m=1m=1;
- if l=3l=3 and r=4r=4 we will have a permutation [1,2][1,2] for m=2m=2;
- if l=2l=2 and r=4r=4 we will have a permutation [3,1,2][3,1,2] for m=3m=3;
- if l=2l=2 and r=5r=5 we will have a permutation [3,1,2,4][3,1,2,4] for m=4m=4;
- if l=1l=1 and r=5r=5 we will have a permutation [5,3,1,2,4][5,3,1,2,4] for m=5m=5.
题意：给出n, n个数, 1~n。问这个排列是不是有x,区间长度为x，之内是1~x的全排列。呃是不是不太明了，看样例。
比如样例3： 1 4 3 2
idx: 1 2 3 4
l=1,r=1, n= 1,可以。
对于2：要想有1,2 idx处是1~4，很明显多了其他的数，不是1~2的全排列。
对于3: 要想有1,2,3，idx处是1~4,4个元素，肯定不行。
对于4：1~4,可以了。
对于这个题，暴力会超时的，所以我们从l,r入手，不断更新l,r，即代码中的,minn，maxx。一定让区间包括1~x，如果需要阔R，就更新maxx,阔L,更新minn：
maxx=max(maxx,idx[i]);
minn=min(minn,idx[i]); //idx记录坐标
如果maxx-minn+1（表示区间内数字的数目）比需要全排列的x(x的全排列就是x个)还大，那肯定不行，所以需要maxx-minn+1==x
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn = 2e5+10; const int inf = 1e9; int idx[maxn]; int main() { int t; cin>>t; while(t--) { int n; cin>>n; int x; for(int i = 1;i<=n;i++ ) { cin>>x; idx[x]=i; } int maxx=-1; int minn = inf; for(int i=1;i<=n;i++) { maxx=max(maxx,idx[i]); minn=min(minn,idx[i]); if((maxx-minn+1)==i) cout<<"1"; else cout<<"0"; } cout<<endl; } }