# Digital Root

## 题目描述

Let $f(n)$ be a sum of digits for positive integer $n$. If $f(n)$ is one-digit number then it is a digital root for $n$ and otherwise digital root of $n$ is equal to digital root of $f(n)$. For example, digital root of $987$ is $6$. Your task is to find digital root for expression $\sum_{i=1}^N \prod_{j=1}^i A_j$.

## 代码

#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

struct IOManager {
template <typename T> inline bool read(T &x) { char c = '\n'; bool flag = false; x = 0; while (~ c && ! isdigit(c = getchar()) && c != '-') ; c == '-' && (flag = true, c = getchar()); if (! ~ c) return false; while (isdigit(c)) x = (x << 3) + (x << 1) + c - '0', c = getchar(); return (flag && (x = -x), true); }
inline bool read(char &c) { c = '\n'; while (~ c && ! (isprint(c = getchar()) && c != ' ')) ; return ~ c; }
inline int read(char s[]) { char c = '\n'; int len = 0; while (~ c && ! (isprint(c = getchar()) && c != ' ')) ; if (! ~ c) return 0; while (isprint(c) && c != ' ') s[len ++] = c, c = getchar(); return (s[len] = '\0', len); }
template <typename T> inline IOManager operator > (T &x) { read(x); return *this; }
template <typename T> inline void write(T x) { if (x < 0) putchar('-'), x = -x; if (x > 9) write(x / 10); putchar(x % 10 + '0'); }
inline void write(char c) { putchar(c); }
inline void write(char s[]) { int pos = 0; while (s[pos] != '\0') putchar(s[pos ++]); }
template <typename T> inline IOManager operator < (T x) { write(x); return *this; }
} io;

struct Solver {
private:
static const int N = 1010;
int n, a[N];
void input() { io > n; }
void process() {
int k; io > k;
for (int i = 1; i <= k; ++ i) io > a[i], a[i] %= 9;
int ans = 9;
for (int i = 1; i <= k; ++ i) {
int mul = 1;
for (int j = 1; j <= i; ++ j) (mul *= a[j]) %= 9;
ans += mul;
}
io < (ans - 1) % 9 + 1 < '\n';
}

public:
void solve() { input(); while (n --) process(); }
} solver;

int main() {
solver.solve();
return 0;
}


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