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 hardness
h(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[/code] [/code]
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 100005
using 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;
}