P2671 [NOIP2015 普及组] 求和

合集 - 模拟(54)

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10.P2671 [NOIP2015 普及组] 求和2024-03-10

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原题链接

题解

得分有哪些条件？颜色相同，下标同奇或同偶，
由此我们对数进行sort分类，把颜色相同的放一起，然后颜色内部按下标小到大排
然后就是很关键的分数计算公式
$(x+z)\times ({\text{number}}_x+{\text{number}}_z).$
可以分解为
${\text{number}}_{x}x+{\text{number}}_{x}z+{\text{number}}_{z}x+{\text{number}}_{z}z$
这下就很明了了，${\text{number}}_{x}x$出现的次数为与x同性的数字的次数，而所有的 ${\text{number}}_{z}x$ 可以和为 $x·Sum-numbe{r}_x$

只要细心，肯定可以想出来的！！

code

#include<bits/stdc++.h>
using namespace std;
const int mod=10007;
struct node
{
    int score,color,x;
}num[100007];

bool cmp(node a,node b)
{
    if(a.color!=b.color) return a.color<b.color;
    return a.x<b.x;
}
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>num[i].score;

    for(int i=1;i<=n;i++)
    {
        cin>>num[i].color;
        num[i].x=i;
    }

    sort(num+1,num+1+n,cmp);
    int j,i=1;
    int ans=0;
    for(;i<=n;)
    {
        int odd=0,even=0,oddsum=0,evensum=0;
        for(j=i;num[j].color==num[i].color;j++)
        {
            if(num[j].x%2==0) even++,evensum+=num[j].score;
            else odd++,oddsum+=num[j].score;
            evensum%=mod;
            oddsum%=mod;
        }
        for(int l=i;l<j;l++)
        {
            if(num[l].x%2==0)
            {
                ans=ans+(num[l].x*((even-1)*num[l].score+evensum-num[l].score));
                ans%=mod;
            }
            else
            {
                ans=ans+(num[l].x*((odd-1)*num[l].score+oddsum-num[l].score));
                ans%=mod;
            }
        }
        i=j;
    }

    cout<<ans;
    return 0;
}