HUTXXXX 数学

这题有个技巧就是保留前缀和后缀级,然后利用公式 a*b MOD c = ( a MOD c * b MOD c ) MOD c。

代码如下:

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define MOD 1000000007
using namespace std;

int N;

long long a[100005], l[100005], r[100005];

int main()
{
    while (scanf("%d", &N), N) {
        for (int i = 1; i <= N; ++i) {
            scanf("%lld", &a[i]);
        }
        l[0] = r[N+1] = 1;
        for (int i = 1; i <= N; ++i) {
            l[i] = (l[i-1]*a[i]) % MOD;
        }
        for (int i = N; i >= 1; --i) {
            r[i] = (r[i+1]*a[i]) % MOD;
        }
        for (int i = 1; i <= N; ++i) {
            printf(i == 1 ? "%lld" : " %lld", (l[i-1]*r[i+1]) % MOD);
        }
        puts("");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/Lyush/p/2594364.html