CF825D:Suitable Replacement(字符串)
来源:互联网 发布:监控软件慧眼下载 编辑:程序博客网 时间:2024/06/08 13:43
You are given two strings s and t consisting of small Latin letters, string s can also contain '?' characters.
Suitability of string s is calculated by following metric:
Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings s, you choose the one with the largest number of non-intersecting occurrences of string t. Suitability is this number of occurrences.
You should replace all '?' characters with small Latin letters in such a way that the suitability of string s is maximal.
The first line contains string s (1 ≤ |s| ≤ 106).
The second line contains string t (1 ≤ |t| ≤ 106).
Print string s with '?' replaced with small Latin letters in such a way that suitability of that string is maximal.
If there are multiple strings with maximal suitability then print any of them.
?aa?ab
baab
??b?za
azbz
abcdabacaba
abcd
In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2.
In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them.
In the third example there are no '?' characters and the suitability of the string is 0.
题意:给出字符串s和t,s可以任意调转字符顺序,要求将问号替换成字符,使得s中包含最多的不相交的t串。
思路:二分答案,设p为s中出现的t数量,二分p即可。
# include <bits/stdc++.h>using namespace std;const int maxn = 1e6+10;char s[maxn], t[maxn], st[maxn];int w=0, cnt=0, T[28]={0}, S[28]={0};int slen, tlen;bool check(int x){ int sum = 0; for(int i=0; i<26; ++i) sum = sum + max(0, T[i]*x-S[i]); return sum <= w;}int main(){ scanf("%s%s",s+1,t+1); slen = strlen(s+1), tlen = strlen(t+1); int p = slen/tlen; for(int i=1; t[i]; ++i) ++T[t[i]-'a']; for(int i=1; s[i]; ++i) { if(s[i]=='?') ++w; else ++S[s[i]-'a']; } int l=0, r=p; while(l<=r) { int m = l+r>>1; if(check(m)) l=m+1; else r=m-1; } for(int i=0; i<26; ++i) { int tmp = max(0, T[i]*r-S[i]); while(tmp--) st[cnt++] = i+'a'; } for(int i=1; i<=slen; ++i) { if(s[i]=='?') { if(cnt) printf("%c",st[--cnt]); else printf("a"); } else printf("%c",s[i]); } return 0;}
- CF825D:Suitable Replacement(字符串)
- codeforces825D-Suitable Replacement
- codeforces 825D Suitable Replacement(字母代替?贪心)
- Codeforces 825 D Suitable Replacement
- Educational Codeforces Round 25 D. Suitable Replacement
- Educational Codeforces Round 25 D. Suitable Replacement
- Codeforces 825D Suitable Replacement【贪心】水题
- 【贪心】codeforces 825D Suitable Replacement
- codeforces825 D. Suitable Replacement 二分答案
- Code Force 825D Suitable Replacement
- Educational Codeforces Round 25 D Suitable Replacement 贪心
- Educational Codeforces Round 25 D. Suitable Replacement【二分】
- Replacement
- Codeforces 135A-Replacement(思维)
- CodeForce 570C Replacement (暴力)
- 精通CSS(parallax&Image replacement)
- codeforces 570-C. Replacement(预处理+规律)
- 用 CSS 实现图片替换文字(Image replacement)
- Hibernate注解
- 第一个静态页面的制作模仿
- HQL查询-分页-条件-连接-过滤使用
- Java中Volatile关键字详解
- android EditText长度监听
- CF825D:Suitable Replacement(字符串)
- 第五章 代数方程求解 和其它符号工具
- 矩阵中最大的两个数
- JMS--java消息中间件(一)
- 快速排序 C++实现
- wampserver apache 500 Internal Server Error(常见问题的解决办法)
- 如何使用apoc 从neo4j导出数据,gephi导入数据
- Jxl导出excel开发实例
- java编写飞行棋