国庆弱校 Carries

B. Carries

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format:  %lld      Java class name:  Main

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frog has  n  integers  a1,a2,…,an , and she wants to add them pairwise.

 

Unfortunately, frog is somehow afraid of carries (进位). She defines \emph{hardness}  h(x,y)  for adding  x  and  y  the number of carries involved in the calculation. For example,  h(1,9)=1,h(1,99)=2 .

 

Find the total hardness adding  n  integers pairwise. In another word, find

∑1≤i<j≤nh(ai,aj)

.

Input

The input consists of multiple tests. For each test:

 

The first line contains  1  integer  n  ( 2≤n≤105 ). The second line contains  n  integers  a1,a2,…,an . ( 0≤ai≤109 ).

Output

For each test, write  1  integer which denotes the total hardness.

Sample Input

2
5 5
10
0 1 2 3 4 5 6 7 8 9

Sample Output

1
20

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

using namespace std;
#define N 100000 + 10

int n, a[N], b[N], x[12];

int main()
{
    x[0] = 1;
    for(int i = 1; i <= 9; i++) x[i] = x[i - 1] * 10;
    while(~scanf("%d", &n))
    {
        for(int i = 0; i < n; i++)
        scanf("%d", &a[i]);

        long long ans = 0;
        for(int k = 0; k <= 9; k++)
        {
            for(int i = 0; i < n; i++)
            b[i] = a[i] % x[k];
            sort(b, b + n);
            for(int i = 0; i < n; i++)
            {
                int tmp = x[k] - b[i];
                int pos = lower_bound(b, b + n, tmp) - b;
                ans += n - pos;
                if(pos <= i) ans--;
            }
        }
        printf("%lld\n", ans >> 1);
    }
    return 0;
}

/*

2
5 5
10
0 1 2 3 4 5 6 7 8 9

*/