bc#27做题笔记

                 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4usingnamespace std;
 5#define mmx 100010
 6#define LL  long long
 7 8struct PER
 9{
10int x,y,dir;
11}p[mmx];
1213LL  a[mmx],A[mmx],ans[mmx];
14int T,n,m;
1516int main()
17{
18     cin>>T;
19while (T--)
20    {
21         cin>>n>>m;
22         memset(A,0,sizeof(A));
23for (int i=1;i<=n-1;i++)
24        {
25//cin>>a[i];               //a[i]:i->i+126             scanf("%d",&a[i]);
27             A[i+1]=A[i]+a[i];        //A[i]:1-->i       A[i]-A[j]: j-->i28        }
2930for (int i=1;i<=m;i++)
31        {
32//cin>>p[i].x>>p[i].y;      //person i : x->y33             scanf("%d%d",&p[i].x,&p[i].y);
34if (p[i].y>p[i].x)  p[i].dir=1; else p[i].dir=2;
35int tx=p[i].x,ty=p[i].y,pos=((i-1)%n)+1;
36if (p[i].dir==1)        //  -->37            {
38if (tx==pos)
39                     ans[i]=A[ty]-A[tx];
40elseif (pos>tx)
41                     ans[i]=A[n]-A[pos]+A[n]+A[ty];
42else43                     ans[i]=A[ty]-A[pos];
44            }
45else//  <--46            {
47if (pos==tx)
48                     ans[i]=A[n]-A[pos]+A[n]-A[ty];
49elseif (pos<tx)
50                     ans[i]=A[n]-A[pos]+A[n]-A[ty];
51else52                     ans[i]=A[n]-A[pos]+A[n]-A[ty];
53            }
54        }
55for (int i=1;i<=m;i++)
56             printf("%I64dn",ans[i]);
57//cout<<ans[i]<<endl;58    }
596061return0;
62 }

                 1 #include <iostream>
 2usingnamespace std;
 3 4 5int main()
 6{
 7     cin>>T;
 8while (T--)
 9    {
10         cin>>n>>m;
11for (int i=0; i<n; i++)
12             scanf("%d",&a[i]);      //Alice13for (int i=n-1;i>=1;i--)
14        {
15             r[i-1]=a[i]/a[i-1];
16        }
17         n--;
18         r[n]=0;
19//r[0...n-1]:text2021longlong ans=0;
22for(int i=1;i<=m;i++)
23        {
24             cin>>p;
25for (int j=0;j<p;j++)
26                 cin>>a[j];
27if (p==1)
28            {
29                 ans+=n+1;
30            }
31else32            {
33for (int j=p-1;j>=1;j--)
34                     b[j-1]=a[j]/a[j-1];
35                 p--;
36                 b[p]=0;
37//b[0..p-1]:pattern38            }
39        }
404142    }
43 }