HDU - 4961 Boring Sum

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5
1 4 2 3 9
0

136
题意:给你一个数组，让你生成两个新的数组，A要求每个数如果能在它的前面找个最近的一个是它倍数的数，那就变成那个数，否则是自己，C是往后找，输出交叉相乘的和
思路：扫描记录因子处理
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef __int64 ll; 
const int MAXN = 100005;

int a[MAXN], b[MAXN], c[MAXN], vis[MAXN];
int n;

int main() {
	while (scanf("%d", &n) == 1) {
		if (n == 0)
			break;
		for (int i = 1; i <= n; i++) 
			scanf("%d", &a[i]); 

		memset(vis, 0, sizeof(vis));
		for (int i = 1; i <= n; i++) {
			if (vis[a[i]])
				b[i] = a[vis[a[i]]];
			else
				b[i] = a[i];
			for (int j = 1; j <= (int)sqrt((double)a[i]+0.5); j++) {
				if (a[i] % j == 0) {
					vis[j] = i;
					vis[a[i] / j] = i;
				}
			}
		}

		memset(vis, 0, sizeof(vis));
		for (int i = n; i >= 1; i--) {
			if (vis[a[i]])
				c[i] = a[vis[a[i]]];
			else
				c[i] = a[i];
			for (int j = 1; j <= (int)sqrt((double)a[i]+0.5); j++) {
				if (a[i] % j == 0) {
					vis[j] = i;
					vis[a[i] / j] = i;
				}
			}
		}

		ll sum = 0;
		for (int i = 1; i <= n; i++) {
			sum += (ll)b[i] * c[i]; 
		}
		printf("%I64d\n", sum); 
	}
	return 0;
}

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HDU - 4961 Boring Sum

标签：des   style   os   io   for   ar   div   amp   

原文地址：http://blog.csdn.net/u011345136/article/details/38689839