1 solutions
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1
#include <iostream> #include <algorithm> using namespace std; const int MAXN = 1e6 + 5; int a[MAXN]; int sum,n; int main(){ cin >> n; for (int i=1;i<=n;i++) cin >> a[i], sum += a[i]; sum /= n; int ans = 0; for (int i=1;i<=n-1;i++) if (a[i]!=sum){ int k = sum - a[i]; //k表示还要拿到几张牌 a[i+1] -= k; //a[i]需要k牌,意味着a[i+1]要拿走k张 ans++; } cout << ans << endl; }
- 1
Information
- ID
- 31
- Time
- 1000ms
- Memory
- 125MiB
- Difficulty
- 2
- Tags
- # Submissions
- 83
- Accepted
- 46
- Uploaded By