题目大意：给你$a_i$，表示第$i$行有$a_i$个空格，你需要确定一个$TAB$宽度，使得剩下的字符最少

题解：做前缀和，发现枚举$TAB$暴力求解是$O(n\log_2n)$的，完全可以过

卡点：无

 

C++ Code：

#include 
     
      #include 
      
       #define maxn 1000010const long long inf = 0x3f3f3f3f3f3f3f3f;int n, M;long long __s[maxn], sum, ans, *s = __s + 1;int main() {	scanf("%d", &n);	for (int i = 1, x; i <= n; i++) {		scanf("%d", &x);		s[x]++;		sum += x;		M = std::max(x, M);	} ans = sum;	for (int i = 1; i <= M; i++) s[i] += s[i - 1];	for (int i = 2; i <= M; i++) {		long long res = 0, j;		for (j = i; j <= M; j += i) {			res += (s[j - 1] - s[j - i - 1]) * (j / i - 1);		}		res += (s[M] - s[j - i - 1]) * (M / i);		ans = std::min(ans, sum - res * (i - 1));	}	printf("%lld\n", ans);	return 0;}