【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=10007;
const int maxn=1000010;
int n,m,ans,ans1;
int c[maxn>>2], c1[maxn>>2], c2[maxn>>2], c3[maxn>>2],c0[maxn>>2],c4[maxn>>2];
int c5[maxn>>2],c6[maxn>>2]; 
struct jgt{
    int a,b;
}t[maxn];


int main(){
    
    scanf("%d%d",&n,&m);
    
    for(int i=1;i<=n;i++)
        scanf("%d",&t[i].b),t[i].b%=mod;
      
    for(int i=1;i<=n;i++)
        scanf("%d",&t[i].a);
    
    for(int i=1;i<=n;i+=2){
        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=1;i<=m;i++) 
    ans = (ans%mod+c1[i]%mod+c2[i]%mod+c4[i]+c6[i])%mod;
    
    memset(c, 0, sizeof(c));memset(c0, 0, sizeof(c0));
    memset(c1, 0, sizeof(c1));memset(c2, 0, sizeof(c2));
    memset(c4, 0, sizeof(c4));memset(c3, 0, sizeof(c3));
    memset(c5, 0, sizeof(c5));memset(c6, 0, sizeof(c6));
    
    for(int i=2;i<=n;i+=2){
        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=1;i<=m;i++) 
    ans = (ans%mod+c1[i]%mod+c2[i]%mod+c4[i]+c6[i])%mod;
    
    printf("%d",(ans+mod)%mod);
    
  }