SCU-4437 Carries(思维题)_mr. f has nn positive integers, a1,a2,…,ana1,a2,…,-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_31759205/article/details/52235487

本文介绍了一种算法，用于解决给定多个整数时，计算所有两两相加过程中产生的进位次数总和的问题。该算法通过将整数按不同位数进行分组，并在每组内进行排序及计算，最终得出所有可能配对相加时的进位总数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Carries

frog has n integers a 1 ,a 2 ,…,a n , and she wants to add them pairwise.

Unfortunately, frog is somehow afraid of carries (进位). She defines hardnessh(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≤10 5 ). The second line contains n integers a 1 ,a 2 ,…,a n . (0≤a i ≤10 9 ).

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

题意:给出n个数,定义h(x,y)是x+y的进位次数,例如h(1,99)=2,因为1+99进了2位,要求求出所有的数相互组合相加进位次数的总和
题解:把每个数字对10,100,...1e9取模,分别把相应取模后的数字存到集合中,在每个集合中,先排序,再枚举第1~n个数,对于每个
数,找到与它相加大于模的那些数,统计这些数的个数全部加起来就是答案

#include<cstdio>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
#include<functional>
#include<iostream>
#include<math.h>
#include<vector>
#include<stdlib.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 5;
LL a[maxn],mod[15];
int main(){
    int n;
    mod[0]=1;
  //  freopen("in.txt","r",stdin);
    for(int i=1;i<10;i++) mod[i]=mod[i-1]*10;
    while(~scanf("%d",&n)){
        vector<int>v[10];
        for(int i=0;i<n;i++) {
            scanf("%lld",&a[i]);
            for(int j=1;j<10;j++)  //对10^j取模
                if(a[i]%mod[j]>0) v[j].push_back(a[i]%mod[j]);//存入相应的集合
        }
        LL ans=0;
        for(int i=1;i<10;i++) {
            sort(v[i].begin(),v[i].end());//从小打到排序,方便二分查找
            for(int j=0;j+1<v[i].size();j++){  //二分查找x+v[i][j]>mod[i]的x的个数
                ans+=v[i].size()-(lower_bound(v[i].begin()+j+1,v[i].end(),mod[i]-v[i][j])-v[i].begin());
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}