#include <cstdio>
2 #include <cstring>
3
4 int color[100005], num[100005], next[100005], n, m, c;
5
6 int main(int argc, char *argv[]){
7 scanf("%d%d", &n, &m);
8 for(int i = 1; i <= n; ++i){
9 scanf("%d", &num[i]);
10 }

11 memset(color, -1, sizeof(color));
12 for(int i = 1; i <= n; ++i){
13 scanf("%d", &c);
14 next[c] = color[c];
15 color[c] = i;
16 }

17 int s = 0;
18 for(int i = 1; i <= m; ++i){
19 for(int z = color[i]; z != -1; z = next[z]){
20 for( int x = next[z]; x != -1; x = next[x]){
21 if((x + z) % 2 == 0 ){
22 s += (x + z) % 10007 * (num[x] + num[z]) % 10007;;
23 }

24 }

25 }

26 }

27 printf("%d", s);
28 return 0;
29 }

~

~

~

~

~

~

~

~

~

~

~

~

~

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,s[100005],y[100005],ans,sb[100005][2],bian[100005][2],sbsbsbsb[100005][2],duang[100005][2];
int main()
{
freopen("p.in","r",stdin);
freopen("p.out","w",stdout);
cin>>n>>m;
for(int i=1;i<=n;i++)
{
scanf("%d",&s[i]);
s[i]=s[i]%10007;
}
for(int i=1;i<=n;i++)
{
scanf("%d",&y[i]);
sb[y[i]][i%2]+=s[i];
sb[y[i]][i%2]%=10007;
bian[y[i]][i%2]+=i;
bian[y[i]][i%2]%=10007;
sbsbsbsb[y[i]][i%2]++;
duang[y[i]][i%2]+=(i*s[i])%10007;
duang[y[i]][i%2]%=10007;
//printf("%d %d %d %d\n",sb[y[i]][i%2],bian[y[i]][i%2],sbsbsbsb[y[i]][i%2],duang[y[i]][i%2]);
}
for(int i=1;i<=m;i++)
{
if(sbsbsbsb[i][0]>=2)
{
ans+=(sbsbsbsb[i][0]-2)%10007*duang[i][0];
ans%=10007;
}
if(sbsbsbsb[i][1]>=2)
{
ans+=(sbsbsbsb[i][1]-2)%10007*duang[i][1];
ans%=10007;
}
}
for(int i=1;i<=m;i++)
{
if(sbsbsbsb[i][0]>=2)
{
ans=ans+(bian[i][0]*sb[i][0])%10007;
ans%=10007;
}
if(sbsbsbsb[i][1]>=2)
{
ans=ans+(bian[i][1]*sb[i][1])%10007;
ans%=10007;
}
//cout<<bian[i][1]<<' '<<sb[i][1]<<endl;
}
cout<<ans;
return 0;
}

#include <cstdio>
#include <iostream>
using namespace std;
long long s[100050][2],ss[100050][2],sss[100050][2],ssss[100050][2],ans=0;
int e[100050],c[100050];
const int q=10007;
int main() {
int n,m,i,tc,fdc;
freopen("sum.in","r",stdin);
freopen("sum.out","w",stdout);
scanf("%d%d",&n,&m);
for (i=1;i<=n;i++) scanf("%d",&e[i]);
for (i=1;i<=n;i++) scanf("%d",&c[i]);

for (i=1;i<=n;i++) {
tc=c[i]; fdc=i %2;
if (ssss[tc][fdc]>0) ans=1LL*ans + 1LL*s[tc][fdc] + 1LL*i*ss[tc][fdc] + 1LL*sss[tc][fdc]*e[i] + 1LL*ssss[tc][fdc]*i*e[i];
s[tc][fdc]=1LL*s[tc][fdc]+1LL*i*e[i];
ss[tc][fdc]=1LL*ss[tc][fdc]+1LL*e[i];
sss[tc][fdc]=1LL*sss[tc][fdc]+1LL*i;
ssss[tc][fdc]=1LL*ssss[tc][fdc]+1LL;
}

cout << ans % q;
}

我就说怎么只对前四个点，仔细一看发现乘法会使int溢出...

车神恒妹儿fromslyz说不开longlong会爆，yveh%%%orz因为我真的爆了，但有什么关系呢，反正我现在AC了

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long n,color[100003],number[100003],num[200003],sum[200003],sc1[200003];
long long m,sc2[200003],ss[200003];
bool p[200003];
int main()
{
scanf("%I64d %I64d\n",&n,&m);
long long i,k,ans=0;
for (i=1;i<=n;++i) scanf("%I64d",&number[i]);
for (i=1;i<=n;++i) scanf("%I64d",&color[i]);
memset(p,0,sizeof(p));
memset(ss,0,sizeof(ss));
memset(sc1,0,sizeof(sc2));
memset(sc2,0,sizeof(sc2));
memset(sum,0,sizeof(sum));
memset(num,0,sizeof(num));
for (i=1;i<=n;++i)
{
k=color[i]*2+i%2;
if (p[k]==0)
{
p[k]=1; num[k]=1;
sc1[k]=number[i]; sc2[k]=i;
ss[k]=sc1[k]*sc2[k];
}
else
{
sum[k]=(sum[k]+((number[i]*i)%10007*num[k])%10007)%10007;
sum[k]=(sum[k]+(number[i]*sc2[k])%10007+(i*sc1[k])%10007)%10007;
sum[k]=(sum[k]+ss[k])%10007;
ss[k]=(ss[k]+(number[i]*i)%10007)%10007;
num[k]++;
sc1[k]=(sc1[k]+number[i])%10007; sc2[k]=(sc2[k]+i)%10007;
}
}
for (i=1;i<=m*2+1;++i)
if (num[i]>1)
ans=(ans+sum[i])%10007;
printf("%I64d\n",ans);
return 0;
}

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
using namespace std;
int i,j,n,m,va[100010],xx,gs[2][100010],xhl[2][100010]; long long zz;
long long lj[4][100010];
long long kal,p;
int main()
{
scanf("%d%d",&n,&m);
for (i=1; i<=n; i++) scanf("%d",&va[i]);
for (i=1; i<=n; i++)
{
scanf("%d",&xx);
kal=lj[i%2+2][xx]; kal%=10007;
zz=xhl[i%2][xx]; zz*=va[i];
kal+=zz; kal%=10007;
zz=lj[i%2][xx]; zz%=10007; zz*=i;
kal+=zz; kal%=10007;
zz=gs[i%2][xx]; zz%=10007; zz*=i; zz%=10007; zz*=va[i];
kal+=zz; kal%=10007;
p+=kal; p=p%10007;
lj[i%2][xx]+=va[i]; gs[i%2][xx]++; xhl[i%2][xx]+=i;
zz=va[i]; zz*=i; lj[i%2+2][xx]+=zz;
}
printf("%I64d",p);
return 0;
}

哎呀，看了题解，码了半天，过了不容易！

type arr1=array[0..100000,0..1]of longint;

var n,m,i,j,tot,r,k,color:longint;sumi,sum,sumnum,count:arr1;

number:array[0..100000]of longint;

begin

readln(n,m);tot:=0;

for i:=1 to n do read(number[i]);

readln;

for i:=1 to n do
begin
read(color);

inc(count[color,i mod 2]);

sumi[color,i mod 2]:=(sumi[color,i mod 2]+i)mod 10007;

sumnum[color,i mod 2]:=(sumnum[color,i mod 2]+number[i])mod 10007;

sum[color,i mod 2]:=(sum[color,i mod 2]+number[i]mod 10007 *i)mod 10007;

end;
for i:=1 to n do

for j:=0 to 1 do

if count[i,j]>1 then

tot:=(tot+sum[i,j]*(count[i,j]-2)+sumi[i,j]*sumnum[i,j])mod 10007;

writeln(tot mod 10007);

end.

其实对于三元组（x，y，z）根本不需要管y，只要x与z同奇或者同偶并且colour[x]==colour[z]（x!=z）就可以满足
纯粹公式，扫一遍就A了···然而之前多写一句废话一直wa

从前往后扫的话，对于编号i，颜色colour[i]的点，那么它与它前面所有和它同奇偶的编号为j，颜色为coulur[i]的三元组的和可以表示为i*（前面所有number[j]的和）+number[i]*（前面所有j的和）+（前面所有number[j]*j）+（前面所有颜色为colour[i]的点的个数）*i*number[i]

相当于把三元组的分数的（x+z）*(number[x]+number[z])拆成number[x]*x+number[x]*z+number[z]*x+number[z]*z

如果前面有n个符合的点，那么就是number[x1]*x1+number[x2]*x2+···+number[xn]*xn///分割线///+（number[x1]+number[x2]+···+number[xn])*z///分割线///+（x1+x2+···+xn)*number[z]///分割线///+n*（number[z]*z）

ps：和楼下的楼下的楼下差不多，可以去看详细解释

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct hehe{
long long w1,w2,w3,w4;
};
hehe a[2][100005];
long long number[100005],colour[100005];
long long ans=0,n,m;
int main(){
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++) scanf("%d",&number[i]);
for (int i=1;i<=n;i++) scanf("%d",&colour[i]);
for (int i=1;i<=n;i++){
ans=(ans+((long long)a[i%2][colour[i]].w1*(long long)number[i])%10007+((long long)a[i%2][colour[i]].w2*(long long)i)%10007+a[i%2][colour[i]].w3+((long long)a[i%2][colour[i]].w4*(long long)(((long long)number[i]*(long long)i)%10007))%10007)%10007;
a[i%2][colour[i]].w1=(a[i%2][colour[i]].w1+i)%10007;
a[i%2][colour[i]].w2=(a[i%2][colour[i]].w2+number[i])%10007;
a[i%2][colour[i]].w3=(a[i%2][colour[i]].w3+((long long)number[i]*(long long)i)%10007)%10007;
a[i%2][colour[i]].w4=(a[i%2][colour[i]].w4+1)%10007;
}
printf("%d",ans%10007);
}

考试时我没把题目写出来，回来后我想到这样写：
∵x，y，z之间的差b为等值
∴可以用等差数列的想法去写
用一条for语句确定x的值，就能得到，y和z
再加上b的值和colour的判断。

评测结果

编译成功

测试数据 #0: Accepted, time = 15 ms, mem = 2128 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 2128 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 2124 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 2120 KiB, score = 10

测试数据 #4: Accepted, time = 250 ms, mem = 2124 KiB, score = 10

测试数据 #5: Accepted, time = 265 ms, mem = 2128 KiB, score = 10

测试数据 #6: TimeLimitExceeded, time = 1031 ms, mem = 2128 KiB, score = 0

测试数据 #7: Accepted, time = 359 ms, mem = 2128 KiB, score = 10

测试数据 #8: Accepted, time = 265 ms, mem = 2124 KiB, score = 10

测试数据 #9: Accepted, time = 312 ms, mem = 2132 KiB, score = 10

TimeLimitExceeded, time = 2497 ms, mem = 2132 KiB, score = 90
代码

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,cl[100010],sum;
struct node
{
int x,n,c;
}st[100010];
bool cmp(node a,node b)
{
if(a.c!=b.c) return a.c<b.c;
else return a.x<b.x;
}
int main()
{
//freopen("p.in","r",stdin);
//freopen("p.out","w",stdout);
cin>>n>>m;
for(int i=1;i<=n;i++) cin>>st[i].n;
for(int i=1;i<=n;i++) cin>>st[i].c;
for(int i=1;i<=n;i++) st[i].x=i;
sort(st+1,st+1+n,cmp);
for(int i=1;i<=n;i++)
{
if(st[i].c!=st[i+1].c) cl[st[i].c]=i;
}
for(int i=1;i<n;i++)
{
for(int j=i+1;j<=cl[st[i].c];j++)
{
if(st[i].c==st[j].c && (st[j].x-st[i].x)%2==0)
{
sum+=((st[i].x+st[j].x)%10007)*((st[i].n+st[j].n)%10007)%10007;
sum%=10007;
}
}
}
cout<<sum;
return 0;
}

这样有90分？？？（手动滑稽