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 $\sum_{i=1}^N \prod_{j=1}^i A_j$.
题意概述
给定一个长度为$N$的数列$A_i$,求$\sum_{i=1}^N \prod_{j=1}^i A_j$的数根。
数据范围:$1 \le N \le 1000, \; 1 \le A_i \le 10^9$。
算法分析
数根有一个有趣的性质:如果$n$等于$0$,那么它的数根就是$0$,否则,它的数根等于$(n-1)\%9+1$。
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; }