CodeForces 621AWet Shark and Odd and Even
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了CodeForces 621AWet Shark and Odd and Even相关的知识,希望对你有一定的参考价值。
Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the n integers, the sum is an even integer 0.
The first line of the input contains one integer, n (1 ≤ n ≤ 100 000). The next line contains n space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Print the maximum possible even sum that can be obtained if we use some of the given integers.
3
1 2 3
6
5
999999999 999999999 999999999 999999999 999999999
3999999996
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999.
题意
求n个数的子序列的和中最大偶数和。
分析
如果n个数总和为偶数,就直接输出,否则减去一个最小的奇数。
代码
#include<stdio.h> long long n,a,minj=1e9+1,sum; int main(){ scanf("%I64d",&n); for(int i=0;i<n;i++) { scanf("%I64d",&a); if(a%2 && a<minj) minj=a; sum+=a; } if(sum%2) sum-=minj; printf("%I64d\n",sum); return 0; }
以上是关于CodeForces 621AWet Shark and Odd and Even的主要内容,如果未能解决你的问题,请参考以下文章
codeforces 621B Wet Shark and Bishops
codeforces 621C Wet Shark and Flowers
CodeForces 621B Wet Shark and Bishops
codeforces 621A Wet Shark and Odd and Even