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最新推荐文章于 2018-11-03 13:27:00 发布

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最新推荐文章于 2018-11-03 13:27:00 发布

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本文链接：https://blog.youkuaiyun.com/qq_37493070/article/details/76718785

本篇介绍一种算法，用于计算给定整数数组中任意两数相加时产生的进位次数总和，并通过示例代码展示了如何利用排序和二分查找优化计算过程。

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Carries

frog has nn integers a1,a2,…,ana1,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 xx and yy the number of carries involved in the calculation. For example, h(1,9)=1,h(1,99)=2h(1,9)=1,h(1,99)=2.

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

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

Input

The input consists of multiple tests. For each test:

The first line contains 11 integer nn (2≤n≤1052≤n≤105). The second line contains nn integers a1,a2,…,ana1,a2,…,an. (0≤ai≤1090≤ai≤109).

Output

For each test, write 11 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

题意：

给你n个数，问其中任意两个数相加能够有多少次进位。

例子：1+99=100，个位进位一次，十位进位一次，就是两次；
50+50 十位进位一次，就是一次。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define LL long long
using namespace std;
LL a[100010];
LL temp[100010];
int main()
{

    LL t;
    while(cin>>t)
    {

        for(int i=0;i<t;i++)
        {
            cin>>a[i];
        }
        LL mod=1;
        LL ans=0;
        for(int i=0;i<9;i++)
        {
            mod*=10;
            for(int j=0;j<t;j++)
            {
                temp[j]=a[j]%mod;
            //  cout<<temp[j]<<"---"<<a[j]<<endl;
            }
            sort(temp,temp+t);
            LL tt;
            for(int j=0;j<t;j++)
            {   tt=t-(lower_bound(temp+1+j,temp+t,mod-temp[j])-temp);
                ans+=tt;
                //      cout<<tt<<endl;
            }
        }       cout<<ans<<endl;

    }


    return 0;
 }