2015 Sichuan Province Contest （Carries）

Carries

Time Limit: 1000ms
Memory Limit: 65536KB
64-bit integer IO format: %lld      Java class name:
Main


Type:
None

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

Source

2015 Sichuan Province Contest

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cmath>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
const double PI = acos(-1.0);
using namespace std;
#define esp  1e-8
const int inf = 99999999;
const int mod = 1000000007;
//freopen("in.txt","r",stdin); //输入重定向,输入数据将从in.txt文件中读取
//freopen("out.txt","w",stdout); //输出重定向,输出数据将保存在out.txt文件中
int a[100005];
int main()
{
int n, i, j;
while (~scanf("%d", &n))
{
long long s = 0;
for (i = 1; i <= n; ++i)
scanf("%d", &a[i]);

for (i = 1e9; i >= 10; i /= 10)//从高位开始每一位位判断是否会进位
{
for (j = 1; j <= n; ++j)
{
a[j] %= i;//提取当前位以下的值
}
sort(a + 1, a + 1 + n);
int l = 1, r = n;
while (l < r)//二分查找
{
if(a[l] + a[r] >= i && l < r)
{
s += r - l;
r --;
}
while (a[l] + a[r] < i && l < r)
{
l++;
}
}
}
printf("%lld\n", s);
}
}