其他数字的积

题目地址："https://www.lanqiao.cn/problems/317/learning/?page=1&first_category_id=1&tags=前缀和&tag_relation=intersection&difficulty=30&difficulty_level=1"

思路一：模拟
直接分类计算就好了，统计一下零的个数cnt，如果cnt>2就全为0，cnt=1就只有为零的那个数字哪里不输出0，cnt=0就正常计算。

include <bits/stdc++.h>

using namespace std;
typedef unsigned long long ull;
const int SIZE = 1e4+5;
int nums[SIZE];
ull sum;
ull cnt;
int main() {
ull N;
cin >> N;
for (int i = 1; i <= N; ++i) {
cin >> nums[i];
if (nums[i]) {
if (!sum)
sum = nums[i];
else {
sum *= nums[i];
}
} else {
cnt++;
}
}

for (int i = 1; i <= N; ++i) {
if (cnt == 0) {
printf("%llu ", sum / nums[i]);
} else if (cnt == 1) {
if (nums[i]) {
printf("0 ");
} else {
printf("%llu ", sum);
}
} else {
printf("0 ");
}
}
}
思路二：前缀积与后缀积
此题目的输出因为i前面那些元素的乘积和i后面那些元素的乘积相乘。只需要记录前缀积和后缀积然后相乘即可，然而此思路效率更低了，除了无需处理0外无任何优势，仅记录：

include

using namespace std;
typedef unsigned long long llu;

llu nums[10005];
llu pre[10005];
llu suf[10005];

int main() {
llu N;
cin >> N;
pre[0] = 1;
for (llu i = 1; i <= N; ++i) {
cin >> nums[i];
pre[i] = pre[i - 1] * nums[i];
}
suf[N + 1] = 1;
for (llu i = N; i > 0; --i) {
suf[i] = suf[i + 1] * nums[i];
}

for (llu i = 1; i <= N; ++i) {
printf("%llu ", pre[i - 1]*suf[i + 1]);
}

}