3 solutions
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1
#include<bits/stdc++.h> using namespace std; const int N=1e5+10; long long a[N],b[N]; int n; int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); cin>>n; for(int i=1;i<=n;i++) cin>>a[i]; b[1]=a[1]; long long tmp1=0,tmp2=0; //tmp1表示b2~bn中所有正数的和 //tmp2表示b2~bn中所有负数取绝对值之后的总和 for(int i=2;i<=n;i++){ b[i]=a[i]-a[i-1]; if(b[i]>0) tmp1=tmp1+b[i]; else tmp2=tmp2+abs(b[i]); } cout<<max(tmp1,tmp2)<<"\n"; cout<<abs(tmp1-tmp2)+1; return 0; }
Information
- ID
- 172
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 6
- Tags
- # Submissions
- 53
- Accepted
- 18
- Uploaded By