var
n,m,i,ans,k:longint;
a:array[0..100001,1..2]of longint;
procedure dfs;
var
i,x,y,z:longint;
begin
for i:=1 to n-2 do begin
k:=0;
while i+k*2<=n do begin
inc(k);
x:=i;
y:=i+k;
z:=i+k*2;
if (a[x,2]=a[z,2]) and (x<>0) then begin
ans:=(ans+(x+z)*(a[x,1]+a[z,1])) mod 10007;
end
end;
end;
end;
begin
readln(n,m);
for i:=1 to n do read(a[i,1]);
for i:=1 to n do read(a[i,2]);
dfs;
writeln(ans);
end.
2 条评论
-
larryzhong LV 9 @ 2016-11-17 21:32:30
40分，过不了。哪位大神指教一下！
#include <iostream>
using namespace std;typedef struct
{
int number;
int color;
} paper;int main()
{
ios::sync_with_stdio(false);
int n, m;
cin>>n>>m;
paper a[n];
for (int i=0; i<n; i++)
cin>>a[i].number;
for (int i=0; i<n; i++)
cin>>a[i].color;
int sum = 0;
for (int x=0; x<n-2; x++)
for (int y=x+1; 2*y-x<n; y++)
{
int z = 2 * y - x;
if (a[x].color == a[z].color)
{
sum += (x+z+2) * (a[x].number+a[z].number);
sum %= 10007;
}
}
cout<<sum<<endl;return 0;
} -
@ 2016-06-10 21:15:55
代码
var n,m,i,sum:longint;
num,color,t,tc,tn,tm:array[1..100000]of Int64;begin
read(n,m);
for i:=1 to n do
read(num[i]);
for i:=1 to n do
read(color[i]);
fillchar(t,sizeof(t),0);
fillchar(tc,sizeof(tc),0);
fillchar(tn,sizeof(tn),0);
fillchar(tm,sizeof(tm),0);
sum:=0;
i:=1;
while i<=n do
begin
inc(t[color[i]]);
inc(tc[color[i]],i*num[i]);
inc(tn[color[i]],num[i]);
inc(tm[color[i]],i);
inc(i,2);
end;
for i:=1 to m do
if t[i]>1 then
sum:=(sum+(t[i]-2)*tc[i]+tn[i]*tm[i])mod 10007;
fillchar(t,sizeof(t),0);
fillchar(tc,sizeof(tc),0);
fillchar(tn,sizeof(tn),0);
fillchar(tm,sizeof(tm),0);
i:=2;
while i<=n do
begin
inc(t[color[i]]);
inc(tc[color[i]],i*num[i]);
inc(tn[color[i]],num[i]);
inc(tm[color[i]],i);
inc(i,2);
end;
for i:=1 to m do
if t[i]>1 then
sum:=(sum+(t[i]-2)*tc[i]+tn[i]*tm[i])mod 10007;
writeln(sum);
end.
- 1
