Digital Root
Problem Description
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 A1*A2*...*AN A1*A2*...*AN-1 ... A1*A2 A1.
Input
Input file consists of few test cases. There is K (1<=K<=5) in the first line of input. Each test case is a line. Positive integer number N is written on the first place of test case (N<=1000). After it there are N positive integer numbers (sequence A). Each of this numbers is non-negative and not more than 109.
Output
Write one line for every test case. On each line write digital root for given expression.
Sample Input
代码语言：javascript复制1
3 2 3 4Sample Output
代码语言：javascript复制5Solution
代码语言：javascript复制#include <bits/stdc .h>
#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4
class BigNum
{
private:
int a[10000];
int len;
public:
BigNum(){ len = 1;memset(a,0,sizeof(a)); }
BigNum(const int);
BigNum(const char*);
BigNum(const BigNum &);
BigNum &operator=(const BigNum &);
friend std::istream& operator>>(std::istream&, BigNum&);
friend std::ostream& operator<<(std::ostream&, BigNum&);
BigNum operator (const BigNum &) const;
BigNum operator-(const BigNum &) const;
BigNum operator*(const BigNum &) const;
BigNum operator/(const int &) const;
BigNum operator^(const int &) const;
int operator%(const int &) const;
bool operator>(const BigNum & T)const;
bool operator>(const int & t)const;
};
BigNum::BigNum(const int b)
{
int c,d = b;
len = 0;
memset(a,0,sizeof(a));
while(d > MAXN)
{
c = d - (d / (MAXN 1)) * (MAXN 1);
d = d / (MAXN 1);
a[len ] = c;
}
a[len ] = d;
}
BigNum::BigNum(const char*s)
{
int t,k,index,l,i;
memset(a,0,sizeof(a));
l=strlen(s);
len=l/DLEN;
if(l%DLEN)
len ;
index=0;
for(i=l-1;i>=0;i-=DLEN)
{
t=0;
k=i-DLEN 1;
if(k<0)
k=0;
for(int j=k;j<=i;j )
t=t*10 s[j]-'0';
a[index ]=t;
}
}
BigNum::BigNum(const BigNum & T) : len(T.len)
{
int i;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i )
a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)
{
int i;
len = n.len;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i )
a[i] = n.a[i];
return *this;
}
std::istream& operator>>(std::istream & in, BigNum & b)
{
char ch[MAXSIZE*4];
int i = -1;
in>>ch;
int l=strlen(ch);
int count=0,sum=0;
for(i=l-1;i>=0;)
{
sum = 0;
int t=1;
for(int j=0;j<4&&i>=0;j ,i--,t*=10)
{
sum =(ch[i]-'0')*t;
}
b.a[count]=sum;
count ;
}
b.len =count ;
return in;
}
std::ostream& operator<<(std::ostream& out, BigNum& b)
{
int i;
out << b.a[b.len - 1];
for(i = b.len - 2 ; i >= 0 ; i--)
{
out.width(DLEN);
out.fill('0');
out << b.a[i];
}
return out;
}
BigNum BigNum::operator (const BigNum & T) const
{
BigNum t(*this);
int i,big;
big = T.len > len ? T.len : len;
for(i = 0 ; i < big ; i )
{
t.a[i] =T.a[i];
if(t.a[i] > MAXN)
{
t.a[i 1] ;
t.a[i] -=MAXN 1;
}
}
if(t.a[big] != 0)
t.len = big 1;
else
t.len = big;
return t;
}
BigNum BigNum::operator-(const BigNum & T) const
{
int i,j,big;
bool flag;
BigNum t1,t2;
if(*this>T)
{
t1=*this;
t2=T;
flag=0;
}
else
{
t1=T;
t2=*this;
flag=1;
}
big=t1.len;
for(i = 0 ; i < big ; i )
{
if(t1.a[i] < t2.a[i])
{
j = i 1;
while(t1.a[j] == 0)
j ;
t1.a[j--]--;
while(j > i)
t1.a[j--] = MAXN;
t1.a[i] = MAXN 1 - t2.a[i];
}
else
t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[t1.len - 1] == 0 && t1.len > 1)
{
t1.len--;
big--;
}
if(flag)
t1.a[big-1]=0-t1.a[big-1];
return t1;
}
BigNum BigNum::operator*(const BigNum & T) const
{
BigNum ret;
int i,j,up;
int temp,temp1;
for(i = 0 ; i < len ; i )
{
up = 0;
for(j = 0 ; j < T.len ; j )
{
temp = a[i] * T.a[j] ret.a[i j] up;
if(temp > MAXN)
{
temp1 = temp - temp / (MAXN 1) * (MAXN 1);
up = temp / (MAXN 1);
ret.a[i j] = temp1;
}
else
{
up = 0;
ret.a[i j] = temp;
}
}
if(up != 0)
ret.a[i j] = up;
}
ret.len = i j;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
BigNum BigNum::operator/(const int & b) const
{
BigNum ret;
int i,down = 0;
for(i = len - 1 ; i >= 0 ; i--)
{
ret.a[i] = (a[i] down * (MAXN 1)) / b;
down = a[i] down * (MAXN 1) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
int BigNum::operator %(const int & b) const
{
int i,d=0;
for (i = len-1; i>=0; i--)
{
d = ((d * (MAXN 1))% b a[i])% b;
}
return d;
}
BigNum BigNum::operator^(const int & n) const
{
BigNum t,ret(1);
int i;
if(n<0)
exit(-1);
if(n==0)
return 1;
if(n==1)
return *this;
int m=n;
while(m>1)
{
t=*this;
for( i=1;i<<1<=m;i<<=1)
{
t=t*t;
}
m-=i;
ret=ret*t;
if(m==1)
ret=ret*(*this);
}
return ret;
}
bool BigNum::operator>(const BigNum & T) const
{
int ln;
if(len > T.len)
return true;
else if(len == T.len)
{
ln = len - 1;
while(a[ln] == T.a[ln] && ln >= 0)
ln--;
if(ln >= 0 && a[ln] > T.a[ln])
return true;
else
return false;
}
else
return false;
}
bool BigNum::operator >(const int & t) const
{
BigNum b(t);
return *this>b;
}
int main() {
std::ios::sync_with_stdio(false);
int k;
std::cin >> k;
while(k--) {
int n;
std::cin >> n;
BigNum total_sum{0};
BigNum prefix_sum{1};
while(n--) {
int num;
std::cin >> num;
prefix_sum = prefix_sum * BigNum{num};
total_sum = total_sum prefix_sum;
}
std::stringstream string_stream{};
string_stream << total_sum;
std::cout.flush();
int digit_sum = 0;
for(auto c : string_stream.str()) {
digit_sum = c - '0';
}
while(static_cast<int>(std::log10(digit_sum)) 1 > 1) {
int new_sum = 0;
while(digit_sum != 0) {
new_sum = digit_sum % 10;
digit_sum /= 10;
}
digit_sum = new_sum;
}
std::cout << digit_sum << std::endl;
}
return 0;
}
