【桶哥的问题——吃桶-简化版】【洛谷p2671】求和

求和=>【链接】

题目相较起_rqy出的要简单很多，来自noip普及组2015

化简这个式子：x+z=2y，故x与z mod 2同余，因此和桶哥的问题——吃桶一样的思路就可以做出来啦qwq：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

const int mod=;
const int maxn=;
int n,m,ans,ans1;
int c[maxn>>], c1[maxn>>], c2[maxn>>], c3[maxn>>],c0[maxn>>],c4[maxn>>];
int c5[maxn>>],c6[maxn>>];
struct jgt{
    int a,b;
}t[maxn];

int main(){

    scanf("%d%d",&n,&m);

    for(int i=;i<=n;i++)
        scanf("%d",&t[i].b),t[i].b%=mod;

    for(int i=;i<=n;i++)
        scanf("%d",&t[i].a);

    for(int i=;i<=n;i+=){
        c1[t[i].a] = (c1[t[i].a]%mod+c[t[i].a]%mod);
        c[t[i].a] = (c[t[i].a]%mod+(i%mod)*(t[i].b%mod))%mod;
        c2[t[i].a] = (c2[t[i].a]%mod+(i%mod)*(c0[t[i].a])%mod)%mod;
        c0[t[i].a] = (c0[t[i].a]%mod+t[i].b)%mod;
        c4[t[i].a] = (c4[t[i].a]%mod+c3[t[i].a]%mod*t[i].b%mod)%mod;
        c3[t[i].a] = c3[t[i].a]%mod+i%mod;
        c6[t[i].a] = (c6[t[i].a]%mod+t[i].b%mod*c5[t[i].a]%mod*i%mod)%mod;
        c5[t[i].a] ++;
    }

    for(int i=;i<=m;i++)
    ans = (ans%mod+c1[i]%mod+c2[i]%mod+c4[i]+c6[i])%mod;

    memset(c, , sizeof(c));memset(c0, , sizeof(c0));
    memset(c1, , sizeof(c1));memset(c2, , sizeof(c2));
    memset(c4, , sizeof(c4));memset(c3, , sizeof(c3));
    memset(c5, , sizeof(c5));memset(c6, , sizeof(c6));

    for(int i=;i<=n;i+=){
        c1[t[i].a] = (c1[t[i].a]%mod+c[t[i].a]%mod);
        c[t[i].a] = (c[t[i].a]%mod+(i%mod)*(t[i].b%mod))%mod;
        c2[t[i].a] = (c2[t[i].a]%mod+(i%mod)*(c0[t[i].a])%mod)%mod;
        c0[t[i].a] = (c0[t[i].a]%mod+t[i].b)%mod;
        c4[t[i].a] = (c4[t[i].a]%mod+c3[t[i].a]%mod*t[i].b%mod)%mod;
        c3[t[i].a] = c3[t[i].a]%mod+i%mod;
        c6[t[i].a] = (c6[t[i].a]%mod+t[i].b%mod*c5[t[i].a]%mod*i%mod)%mod;
        c5[t[i].a] ++;
    }

    for(int i=;i<=m;i++)
    ans = (ans%mod+c1[i]%mod+c2[i]%mod+c4[i]+c6[i])%mod;

    printf("%d",(ans+mod)%mod);

  }