Given a and n, calculate a + aa + aaa +... Sum of a (n a) (Big Data Processing)

Description:

Given a and n, calculate the sum of a + aa + aaa + a... a (n.

Input:

There are multiple groups of test data. Input a, n (1 <= a <= 9,1 <= n <= 100 ).

Output:

Output results for each group of inputs.

Sample input:

1 10

Sample output:

1234567900

From the question, we can see that when a = 9, n = 100, the number of an int type is no less than 100 bits, so we cannot use the common method to find it, the following describes my solution. I declare a vector v to store a + aa + aaa +... the sum of a (n a), temp is used to store... for a (n a), it is added separately from a bit to a high bit, and hight is used to store carry.

# Include <iostream> # include <vector> using namespace std; int main () {int a, n; int sum = 0; vector <int> v; vector <int> temp; vector <int>: iterator it; int hight = 0; // store carry while (cin> a> n) {v. clear (); temp. clear (); v. push_back (a); temp. push_back (a); for (int I = 2; I <= n; I ++) {temp. push_back (a); int j = temp. size ()-1; int k = v. size ()-1; hight = 0; sum = 0; while (k> = 0 & j> = 0) {// sum = temp from low to high [j] + V [k] + hight; hight = 0; if (sum> 9) {hight = sum/10; // returns the bitwise} v [k] = sum % 10; k --; j --;} // if (hight> 0) {while (j> = 0) {// you may need to add more numbers than the total number of digits, for example, 9 + 99; sum = temp [j] + hight; hight = 0; if (sum> 9) {hight = sum/10;} v. insert (v. begin (), sum % 10); j --;} if (hight> 0) {// if there is a carry, put it to the highest digit v. insert (v. begin (), hight);} //} for (it = v. begin (); it! = V. end (); it ++) {cout <* it;} cout <endl;} return 0 ;}

Result: a = 9 n = 200, a + aa + aaa + a... a = 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110910

The following is a simple solution: directly simulate the primary addition from the single digit to add, the carry, and then store it in a stack. Finally, the output of the stack is finished. Code:

#include <cstdio>#include <stack>using namespace std;int main(){//    freopen("1.txt", "r", stdin);    int a, n, i, t, c;    while(~scanf("%d %d", &a, &n))    {        stack<int> S;        for(c=0,i=1; i<=n; i++)        {            t = (n-i+1)*a;            S.push((t+c)%10);            c = (t+c)/10;        }        if(c>0)            S.push(c);        while(!S.empty())        {            printf("%d", S.top());            S.pop();        }        printf("\n");    }    return 0;}

However, this disadvantage is that when n is large, t = (n-I + 1) * a will overflow, and this program runs quickly, the above code runs slowly when n is large. But it does not overflow.