弱校联萌十一大决战之强力热身 B题

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

我用字符串做的！但是一直wa，和别人ac的代码一直对比测试数据，答案都对得上！不知道哪里错了！若有人看到，希望能得到指教！

别人的ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<algorithm>
#include<vector>
#include<bitset>
using namespace std;
const int mmax = 100010;
typedef long long LL;
int a[mmax];
int b[mmax];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int tmp=1;
        LL ans=0;
        for(int i=0;i<9;i++)
        {
            tmp*=10;
            for(int j=0;j<n;j++)
                b[j]=a[j]%tmp;
            sort(b,b+n);
            int r=n;
            for(int j=0;j<n;j++)
            {
                while(  r && b[j]+b[r-1]>=tmp)
                    r--;
                if(r<=j)
                    ans+=(n-r-1);
                else
                    ans+=(n-r);
            }
        }
        ans/=2;
        printf("%lld\n",ans);
    }
    return 0;
}

我自己wa的代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
typedef long long ll;
using namespace std;
int main()
{
  ll n;
  while(~scanf("%lld",&n))
  {
    ll cnt=0;
    char ch[13];
    char ch1[13];
    char ch2[13];
    memset(ch,0,sizeof(ch));
    memset(ch1,0,sizeof(ch1));
    memset(ch2,0,sizeof(ch2));
    ll arr[100005];
    for(ll i=0;i<n;++i)
      cin>>arr[i];
    sort(arr,arr+n);
    for(ll j=n-1;j>=0;--j)
    {
      if(j>=1&&arr[j]+arr[j-1]<10)
        break;
      for(ll i=j-1;i>=0;--i)
      {
        if(arr[i]+arr[j]<10)
          break;
        sprintf(ch1,"%lld",arr[i]);
        sprintf(ch2,"%lld",arr[j]);
        ll lenth1=strlen(ch1);
        ll lenth2=strlen(ch2);
        ll min_lenth=min(lenth1,lenth2);
        ll max_lenth=max(lenth1,lenth2);
        ll t=min_lenth-1,k=max_lenth-1;
        for(;t>=0;--t,--k)
        {
          ll ans=ch1[t]-'0'+ch2[k]-'0';
        //  cout<<ch1[t]-'0'<<" "<<ch2[k]-'0'<<endl;
          if(ans>=10)
            {
              ch2[k-1]+=1;
          //    cout<<"ch"<<ch2[k-1]-'0'<<" "<<k-1<<endl;
              cnt++;
            }
        }
     //   cout<<k<<endl;
        if(min_lenth!=max_lenth&&ch2[k]-'0'>=10)
        {
          //cout<<"sdfsdf"<<ch2[k]-'0'<<endl;
          cnt++;
        }
      }
    }
    cout<<cnt<<endl;
  }
  return 0;
}