BZOJ1045 [HAOI2008] 糖果传递 【数学】
题解：
跟均分纸牌差不多
# include <stdio.h>
# include <algorithm>
using namespace std;
int n, a[10000010], p[10000010];
long long sum;
/*
p[i] = p[i-1] + sum - a[i]
p[1] = p[n] + sum - a[1]
p[n] = sum - a[1] - p[1];
sigma (p[1] + p[2] +...+p[n])
*/
int main() {
scanf("%d", &n);
for (int i=1; i<=n; ++i) scanf("%d", &a[i]), sum+=a[i];
sum=sum/n;
for (int i=2; i<=n; ++i) p[i] = p[i-1] + sum - a[i];
sort(p+1, p+n+1);
int mid = p[n/2];
long long ans=0;
for (int i=1; i<=n; ++i)
ans += abs(mid - p[i]);
printf("%lld\n", ans);
return 0;
}
2023年4月14日 23:20
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