C. Training Before the Olympiad
1.C. Insert and Equalize2.C. Removal of Unattractive Pairs3.D. Jumping Through Segments4.E. Good Triples5.F. Shift and Reverse6.D. Yet Another Monster Fight7.A. Constructive Problems8.C. Game with Multiset9.A. Rating Increase10.B. Swap and Delete11.A. Problemsolving Log12.B. Preparing for the Contest13.C. Quests14.D. Three Activities15.E2. Game with Marbles (Hard Version)16.cf刷题有感17.A. Anonymous Informant18.A. Forked!19.B. Make Almost Equal With Mod20.C. Heavy Intervals21.D. Split Plus K22.A. 202323.B. Two Divisors
24.C. Training Before the Olympiad
25.D. Mathematical Problem26.F. Greetings27.C. Partitioning the Array28.G. Bicycles29.E. Eat the Chip30.G. Lights31.D. Array Repetition32.D. Berserk Monsters33.E. Increasing Subsequences34.D. Very Different Array35.G. Mischievous Shooter36.B. Plus-Minus Split37.B. A Balanced Problemset38.C. Did We Get Everything Covered39.D. Find the Different Ones!40.C. Grouping Increases41.D. Good Trip42.C. Physical Education Lesson43.E. Final Countdown44.D. Divisible Pairs45.G. Vlad and Trouble at MIT46.A. Brick Wall47.B. Minimize Inversions48.C. XOR-distance49.A. Moving Chips50.B. Monsters Attack!51.C. Find B52.D. Slimes53.C. Turtle Fingers: Count the Values of k54.D. Turtle Tenacity: Continual Mods55.D. Vlad and Division56.C. LR-remainders57.D. Lonely Mountain Dungeons题解
事实1.两个数合并后的数一定是偶数
事实2.如果没有发生奇数和偶数合并,那么最终的结果一定是所有数的和
事实3.每发生一次奇数和偶数合并,最后的结果会减一
总结
综上所述,Masha会尽量选择同奇或同偶合并,但在有同奇的情况下,会优先选择同奇,因为合并会产生偶数,且Olya需要用到奇数与偶数合并
代码
#include<bits/stdc++.h>
using namespace std;
long long sum[100005]={0},a[100005]={0};
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
for(int i=1;i<=n;i++){cin>>a[i];sum[i]=sum[i-1]+a[i];}
cout<<a[1];
long long js=a[1]%2;
for(int i=2;i<=n;i++)
{
js+=a[i]%2;
cout<<" ";
if(js%3%2)cout<<sum[i]-js/3-1;
else cout<<sum[i]-js/3;
}
puts("");
}
return 0;
}
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