[Uva1642]魔法Gcd（数论）

Description

给定n个数，某个连续区间[L,R]的收益为\(gcd(A_l,A_{l+1},A_{l+2}...A_r)*(r-l+1)\)，

求收益最大的区间的收益值

\(1 \leq n \leq 50000,A_i<=10^9\)

Solution

设f[i][j]为区间[i,j]的gcd，那么就有\(f[i][j]=gcd(f[i][j-1],A_i)\) ，

由此可以固定右端点算出每个区间的gcd，同时更新Ans

用一个数组储存当前所有gcd的值，

如果有相同的gcd，与上一个区间合并即可，否则增加一个新的gcd的值

Code

#include <cstdio>
#include <algorithm>
#define N 500010
using namespace std;

int n, A[N], tot, l[N];
long long Ans;

inline int read() {
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') {if (ch == '-')f = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
	return x * f;
}

int gcd(int a, int b) {
	return (b == 0) ? a : gcd(b, a % b);
}

int main() {
	n = read();
	for (int k, i = 1; i <= n; ++i) {
		A[++tot] = read();
		l[tot] = 1, k = 0;
		for (int j = 1; j <= tot; ++j) {
			A[j] = gcd(A[j], A[tot]);
			if (A[j] == A[k]) l[k] += l[j];
			else A[++k] = A[j], l[k] = l[j];
		}
		int s = 0; tot = k;
		for (int j = tot; j; --j) {
			s += l[j];
			Ans = max(Ans, 1ll * A[j] * s);
		}
	}
	printf("%lld\n", Ans);
	return 0;
}