CodeForces - 1251B - Binary Palindromes(思维)
作者:互联网
题目链接:https://vjudge.net/problem/CodeForces-1251B
题目大意:给你若干个01串,你可以交换任意一对字符,可以在同一个串,也可以在不同串
因为题目的限制很自由,所有我们只要判断奇偶就行了.注释应该写的很详细了
#include<set> #include<map> #include<list> #include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl '\n' #define rtl rt<<1 #define rtr rt<<1|1 #define lson rt<<1, l, mid #define rson rt<<1|1, mid+1, r #define maxx(a, b) (a > b ? a : b) #define minn(a, b) (a < b ? a : b) #define zero(a) memset(a, 0, sizeof(a)) #define INF(a) memset(a, 0x3f, sizeof(a)) #define IOS ios::sync_with_stdio(false) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; typedef pair<char, int> P2; const double pi = acos(-1.0); const double eps = 1e-7; const ll MOD = 1000000007LL; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const double EULC = 0.5772156649015328; const int NIL = -1; template<typename T> void read(T &x){ x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f; } const int maxn = 1e3+10; int main(void) { int t; cin >> t; while(t--) { int n, n0 = 0, n1 = 0, odds = 0; string s; cin >> n; for (int i = 0; i<n; ++i) { cin >> s; int size = s.size(); if (size&1) ++odds; //统计奇数串数目 for (auto ch : s) { //统计0和1的数目 if (ch=='1') ++n1; else if (ch=='0') ++n0; } } if (odds==1 && n1&1 && n0&1) cout << n-1 << endl; //如果只有1个奇数串,则只能有一种数字是奇数 else if (odds&1 && ((~n0&1 && ~n1&1) || (n0&1 && n1&1))) cout << n-1 << endl; //如果奇数串的数量是奇数而且大于1,那么只有在只有1种字符数量为奇数的时候可行 else if (!odds && ((n1&1)||(n0&1))) cout << n-1 << endl; //如果没有奇数串那么两种数字都不能出现奇数 else if (~odds&1 && ((n1&1 && ~n0&1) || (~n1&1 && n0&1))) cout << n-1 << endl; //如果奇数串的数量不为1且大于0, 那么只要两种数字的数量都是偶数就行 else cout << n << endl; } return 0; }
标签:Binary,typedef,ch,const,int,CodeForces,Palindromes,include,define 来源: https://www.cnblogs.com/shuitiangong/p/12437407.html