HDU_5504 GT and sequence
GT and sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1392 Accepted Submission(s): 322
N
integers.
You should choose some numbers(at least one),and make the product of them as big as possible.
It guaranteed that **the absolute value of** any product of the numbers you choose in the initial sequence will not bigger than
2
T
(test numbers).
For each test,in the first line there is a number N
in the next line there are N
numbers.
1≤T≤1000
1≤N≤62
You'd better print the enter in the last line when you hack others.
You'd better not print space in the last of each line when you hack others.
1
3
1 2 3
6水题一枚!可是坑了我两三次,结果没注意输入的数可以是long long 型wa了好几次~~//题意:给你若干个整数,让你挑选若干数字使得其乘积最大
//算法分析:将这些数字都进行分类,正数、负数、零分别放在不同的一个数组中,然后再进行讨论!讨论负数时候,若为奇数个,去掉最大的那个,留下的相乘即可!偶数个直接相乘 #include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
using namespace std; typedef long long int LL ; int main() {
int t;
cin >> t;
while (t --) {
int n;
cin >> n;
LL a[100];
LL b[100], c[100];
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(c, 0, sizeof(c));
int num1 = 0, num2 = 0;
int flag = 0;
for (int i = 0; i<n; i++) {
cin >> a[i];
if (a[i] > 0)
b[num1++] = a[i];
else if (a[i] < 0)
c[num2++] = a[i];
else
flag ++ ;
}
sort(c, c+num2);
LL res = 1;
if (num1 == 0) {
if (num2 == 0)
cout << 0<< endl;
else if(num2 == 1) {
if (flag == 0)
cout << c[0] << endl;
else
cout << 0<< endl;
}
else {
if (num2 %2 == 0) {
for (int i = 0; i<num2; i++)
res *= c[i];
}
else {
for (int i = 0; i<num2-1; i++)
res *= c[i];
}
cout << res << endl;
}
}
else {
for (int i = 0; i<num1; i++)
res *= b[i];
if (num2 %2 == 0) {
for (int i = 0; i<num2; i++)
res *= c[i];
}
else {
for (int i = 0; i<num2-1; i++)
res *= c[i];
}
cout << res << endl;
}
} return 0 ;
}有坑跳才会进步!
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