BNUOJ 弱校联萌十一大决战之强力热身B Carries(二分)_carries frog has n integers a1,a2,…,an, and she wa-优快云博客

本文链接：https://blog.youkuaiyun.com/Grit_ICPC/article/details/48848127

本文探讨了在处理大量整数相加时如何减少进位操作的策略，并通过实例展示了如何利用二分查找来降低计算的复杂度。通过具体代码实现，文章详细解释了算法的实现过程和效率提升。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

\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

思路:一对一对的枚举铁定会超时.所以首先要解决这个问题.二分的时间复杂度为logn可以.

那么,还要解决进位的问题.b[j]=a[i]%tmp^k;

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<map>
#include<algorithm>
#include<queue>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
ll a[100000],b[100000];
int main()
{
    long long  n,m,i,j,k;
    while(~scanf("%lld",&n))
    {
        for(i=0; i<n; i++)
            scanf("%lld",&a[i]);
        ll tmp=1;
        ll ans=0;
        for(i=0; i<9; i++) //最多为10^9
        {
            tmp*=10;
            for(j=0; j<n; j++)
                b[j]=a[j]%tmp;
            sort(b,b+n);//为二分先排序,并且不影响进位的次数
            for(j=0; j<n; j++)
            {
                ll l=j+1,r=n;
                while(l<r)
                {
                    int mid=(l+r)>>1;
                    if(b[mid]+b[j]>=tmp)
                        r=mid;
                    else
                        l=mid+1;
                }
                ans+=n-r;
            }
        }
        printf("%lld\n",ans);
        return 0;
    }
}