【思维】SCU 4437 Carries

4437: Carries

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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) for adding xx and yy the number of carries involved in the calculation. For example, h(1,9)=1,h(1,99)=2. Find the total hardness adding nn 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 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：给定n个数的序列，问这个序列任意两个数相加需要进的位数的和是多少；

思路：

1.给定两个数若这两个数的第k位相加会进位，则满足：(a%10^k+b%10^k)>=10^k；

2.按位计算贡献（进位次数）；

3.题目要求是每两个只选择一次，但排序后并不影响，并不会重复枚举两个数；

样例：

  97      56

  56      97

123    123  //更换顺序后不会影响统计贡献（进位次数）；

代码：

#include
#include
using namespace std;
typedef long long ll;

const int maxn=100005;
int a[maxn],b[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);

        int mod=1;
        ll ans=0;
        for(int i=1;i<=9;i++)
        {
            mod*=10;
            for(int i=1;i<=n;i++)
                b[i]=a[i]%mod;
            sort(b+1,b+n+1);
            for(int i=1;i