2016 Multi-University Training Contest 2 1011 Keep On Movin (简单数学)
2016-07-22 15:11
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题目链接
所以这题的回文串个数应该是min(1,num_of_odd) ,最短的回文串长度为(even/2/odd)∗2+1
简单题意
有n种不同的字符,每种有ai个,用所有的字符来构造回文串,使最短的回文串长度最长,并输出该长度;思路
很简单,如果某种字符为奇数个,就要有取出来一个放在回文串中间,剩下的放在回文串两侧,偶数个就直接分配在回文串两侧。所以这题的回文串个数应该是min(1,num_of_odd) ,最短的回文串长度为(even/2/odd)∗2+1
代码
#include <bits/stdc++.h> using namespace std; int main(){ int T; cin >> T; while(T --){ int n; int odd = 0,en = 0; scanf("%d", &n); for(int i = 0 ; i < n ; i ++){ int tt; scanf("%d", &tt); if(tt %2){odd ++;en += tt-1;} else en+= tt; } if(odd < 1) printf("%d\n",en); else printf("%d\n",(en/2/odd)*2 + 1); } }
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