2015四川省acm B题

Carries

frog has n n integers a 1 ,a 2 ,…,a n  a1,a2,…,an, and she wants to add them pairwise.

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

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

∑ 1≤i<j≤n h(a i ,a j ) ∑1≤i<j≤nh(ai,aj)

.

Input

The input consists of multiple tests. For each test:

The first line contains 1 1 integer n n (2≤n≤10 5  2≤n≤105). The second line contains n n integers a 1 ,a 2 ,…,a n  a1,a2,…,an. (0≤a i ≤10 9  0≤ai≤109).

Output

For each test, write 1 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
原题链接： http://acm.scu.edu.cn/soj/problem.action?id=4437

题意：给出几组测试数据，要求分别输出每组数的所有元素相加会发生多少次进位，输出进位数。

方法：分别求每位上的的进位数，直接快速排序，然后两个指针扫描。

代码如下：

#include<iostream>#include<algorithm>#include<cmath>#define maxn 100005using namespace std;long long int mergesort(int a[],int begin,int end) {     int b[maxn];     long long num;     long long count=0;     int n=end+1,p;     int q,r;     for(int k=1;;k++)     {         p=0,q=0,r=end;         num=pow(10,k);         for(int i=0;i<n;i++)         {             b[i]=a[i]%num;             if(b[i]!=a[i])             {                 p=1;             }         }         sort(b,b+n);         while(r!=q)         {             if(b[q]+b[r]>=num)             {                 count+=r-q;                 r--;             }             else              {                 q++;             }         }         if(p==0)         break;     }     return count; }int main(void) {     int n,a[maxn];     while(cin>>n)     {         for(int i=0;i<n;i++)         cin>>a[i];         cout<<mergesort(a,0,n-1)<<endl;     }     return 0; }