2015弱校联盟(1) - B. Carries

B. Carries
Time Limit: 1000ms
Memory Limit: 65536KB

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≤n h(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

将其对10,100,1000……进行取余的结果进行排序,如果两个数的和大于他们的取余数,则有进位

#include <bits/stdc++.h>
#define LL long long
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;

const int Max = 1e5+100;

int a[Max];
int b[Max];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int ans=1;
        LL num=0;
        for(int i=1;i<=9;i++)
        {
            ans*=10;
            for(int k=0;k<n;k++)
            b[k]=a[k]%ans;
            sort(b,b+n);
            int j=0;
            for(int k=n-1;k>=0;k--)
            {
                for(;j<n;j++)
                {
                    if(b[k]+b[j]>=ans)
                    {
                        break;
                    }
                }
                if(j<=k)
                {
                    num--;
                }
                num+=(n-j);
            }

        }
        printf("%lld\n",num/2);
    }
    return 0;
}