标签:long stack sig eof name log typedef force list
题意:给定两个串(长度范围为 1 ~ 2*10^5,两串长度不一定相同,由大、小写字母构成),需要让两串中完全相同字母匹配的尽量多,在此前提下,再让同一字母但大小写不同的对数尽量多。
先尽量多的完全相同匹配,然后再尽量多的匹配大小写不同的字母即可
#include<cstdio> #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<deque> #include<queue> #include<stack> #include<list> typedef long long ll; typedef unsigned long long llu; const int MAXN = 100 + 10; const int MAXT = 1000000 + 10; const int INF = 0x7f7f7f7f; const double pi = acos(-1.0); const double EPS = 1e-6; using namespace std; char s1[MAXT], s2[MAXT], s[MAXT], vis[MAXT]; map<int, int> u; int main(){ memset(vis, 0, sizeof vis); scanf("%s%s", s1, s2); for(int i = 0; s1[i] != ‘\0‘; ++i){ if(isupper(s1[i])) ++u[s1[i] - ‘A‘ + 26]; else ++u[s1[i] - ‘a‘]; } int ans1 = 0, ans2 = 0; for(int i = 0; s2[i] != ‘\0‘; ++i) if(isupper(s2[i])){ if(u[s2[i] - ‘A‘ + 26] >= 1){ --u[s2[i] - ‘A‘ + 26]; ++ans1; vis[i] = 1; } } else{ if(u[s2[i] - ‘a‘] >= 1){ --u[s2[i] - ‘a‘]; ++ans1; vis[i] = 1; } } for(int i = 0; s2[i] != ‘\0‘; ++i){ if(vis[i] == 1) continue; if(isupper(s2[i])){ if(u[s2[i] - ‘A‘]){ --u[s2[i] - ‘A‘]; ++ans2; } } else{ if(u[s2[i] - ‘a‘ + 26]){ --u[s2[i] - ‘a‘ + 26]; ++ans2; } } } printf("%d %d\n", ans1, ans2); return 0; }
CodeForces 518B - Han Solo and Lazer Gun(模拟)
标签:long stack sig eof name log typedef force list
原文地址:http://www.cnblogs.com/tyty-TianTengtt/p/5996053.html