SCU 4437 Carries

最新推荐文章于 2019-08-14 16:51:10 发布

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本文介绍了一种算法，用于计算给定整数集合中任意两数相加时产生的进位总数。通过逐位比较并利用排序与二分查找的方法，有效地解决了这一问题。

Carries

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 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

\sum 1 \leq i < j \leq n h (a i, a j)

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

题解：
判断两个数相加的第i位是否需要进位只需判断
x%(i^10)+y%(i^10)与i^10的关系即可。
若x%(i^10)+y%(i^10)>=i^10成立，则当z%(i^10)>=y%(i^10)时也一定成立
代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int a[100004],b[100004];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)scanf("%d",&a[i]);
        int mod=1;
        ll ans=0;
        for(int i=1; i<=9; i++)
        {
            mod*=10;
            for(int j=0; j<n; j++)b[j]=a[j]%mod;
            sort(b,b+n);
            for(int j=0; j<n; j++)
            {
                int pos=lower_bound(b+j+1,b+n,mod-b[j])-b;
                ans=ans+n-pos;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}