T1:

Explanation:

60 PTS Backpack + program output during the exam... (Expected score 50?
Is it stable or good?
100pts:
Find out the prefix sum, first module n out, sweep to see if there is a prefix sum is 0, if there is a prefix sum is 0, then directly output 1~x (prefix sum is 0)
If the sum of prefixes is zero, there are N prefixes and the range of values is 1-N-1, that is to say, at least two prefixes and the sum of them are the same.
What's the significance of this? If sum [a] = sum[b] (a < = b), then sum[b] sum[a]=0
What does that mean? That's not to say a+1~b comes up with a n n, just output it directly.
This is urinary... It's very similar to the C question of CF.

Code:

60pts

#include <cstdio>
#define N 1000005
using namespace std;
struct hh{int x,id;}a[N];
bool f[N];int tot,ans[N];
inline int in()
{
    char ch=getchar();int x=0,f=1;
    while (ch>'9'|| ch<'0') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
void find(int z,int up)
{
    if (z==0) return;
    for (int i=up;i>=1;i--)
      if (f[z-a[i].x])
      {
        ans[++tot]=a[i].id;
        find(z-a[i].x,i-1);
        break;
      }
}
int main()
{
    freopen("set.in","r",stdin);
    freopen("set.out","w",stdout);
    int n,i,j;
    n=in();
    for (i=1;i<=n;i++) {a[i].x=in(); a[i].id=i;}
    for (i=1;i<=n;i++)
    {
        a[i].x%=n;
        if (a[i].x==0) {printf("1\n%d",a[i].id);return 0;}
     } 
    bool vv=0;f[0]=1;
    for (i=1;i<=n;i++)
    {
        for (j=n;j>=a[i].x;j--)
        {
            f[j]=f[j] || f[j-a[i].x];
            if (f[n]){vv=1;break;} 
        }
        if (vv) break;
    }
    if (!vv){printf("-1");return 0;}
    else 
    {
        find(n,i);
        printf("%d\n",tot);
        for (i=1;i<tot;i++) printf("%d ",ans[i]);
        printf("%d",ans[tot]);
    }    
}

100pts

#include <cstdio>
#include <algorithm>
#define N 1000005
using namespace std;
int sum[N],a[N],c[N];
inline int in()
{
    char ch=getchar();int x=0,f=1;
    while (ch>'9'|| ch<'0') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int main()
{
    freopen("set.in","r",stdin);
    freopen("set.out","w",stdout);
    int n,i,j;
    n=in();
    for (i=1;i<=n;i++) 
    {
        a[i]=in();
        a[i]%=n;sum[i]=(a[i]+sum[i-1])%n;
    }
    for (i=1;i<=n;i++)
      if (!sum[i]) 
      {
        printf("%d\n",i);
        for (j=1;j<=i;j++)
          printf("%d ",j);
        return 0;
      } 
    for (i=1;i<=n;i++)
    {
        if (c[sum[i]])
        {
            printf("%d\n",i-c[sum[i]]);
            for (j=c[sum[i]]+1;j<=i;j++)
              printf("%d ",j);
            return 0;
        }
        c[sum[i]]=i;
    }
    printf("%d",-1);
}

T2:

Explanation:

This question is the original one...
That's to find the modulus of n numbers. If there's no modulus, it's zero.
If there is a number, it is a number * 2 - the total number - 1
Because the number that can be paired is 2*num (the number of modes) - n (the total number) - 1, that's it? It's to put another mode number on the first place.

Code:

#include <cstdio>
#include <algorithm>
#define N 10005
using namespace std;
int cont[N],x[N],y[N],z[N],stack[N],cnt;
inline int in()
{
    char ch=getchar();int x=0,f=1;
    while (ch>'9'|| ch<'0') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int main()
{
    freopen("read.in","r",stdin);
    freopen("read.out","w",stdout);
    int n=0,i,j,m,k;
    m=in();k=in();
    for (i=1;i<=m;i++) cont[i]=in();
    for (i=1;i<=m;i++) x[i]=in();
    for (i=1;i<=m;i++) y[i]=in();
    for (i=1;i<=m;i++) z[i]=in();
    int s=(1<<k)-1;int a=-1,now=0;
    for (i=1;i<=m;i++)
    {
        if (!now) a=x[i];
        if (x[i]!=a) now--;else now++;
        long long last=x[i];
        for (int j=1;j<cont[i];j++)
        {
            last=(last*y[i]+z[i])&s;
            if (!now) a=last;
            if (last!=a) now--;else now++;
        }
    }
    now=0;
    for (i=1;i<=m;i++)
    {
        ++n;
        if (x[i]==a) now++;
        long long last=x[i];
        for (int j=1;j<cont[i];j++)
        {
            last=(last*y[i]+z[i])&s;
            ++n;
            if (last==a) now++;
        }
    }
    now=now*2-n;
    if (now<0) printf("0");else printf("%d",now-1);
}

Make complaints:

Just finished the exam, loser: Ah, the original topic of T2
Meow Meow Meow: What's the problem?
loser: bzoj2456
Meow and Meow (Enter bzoj excitedly): 2-4-5-6
Meow meow meow:...... Meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow meow.
Meow meow: This question... I've done it.
Then take a closer look. woc is really the topic. It just changed shape.
... Goodbye,.jpg.

Is the spj of T1 written?
Then there are four approaches to T1:
Output - 1; Expected score 10; Actual score 0
n^2dp; expected score 50; actual score 60
Scientific problem solving; Expected score 100; Actual score 100
Nothing output; expected score 0; actual score 100
hahaha