A

Problem description

给出数a,b,n,在a后加上1个数,并使加后的数是b的倍数.输出操作n次后的数.

Data Limit：1?≤?a,?b,?n?≤?105  Time Limit: 1s

Solution

易得只要第一次成功,剩下加0,就可以了

Code

#include<cstdio>
long long a,b,n,i,j;
long long ans[500000];bool bo=false;
int main()
{
    scanf("%lld%lld%lld",&a,&b,&n);//a=a%b;
    for (i=0;i<=9;i++)
    {
        if ((a*10+i)%b==0)
        {
            a=a*10+i;bo=true;
            break;
        }
    }
    if (bo)
    {
        printf("%lld",a);
    for (i=1;i<=n-1;i++)printf("%d",0);
    }else printf("%d",-1);
    return 0;
}

B

Problem description

输入=一个字符串,输出出现最多的合法日期.

Data Limit：|s|<=10e5  Time Limit: 1s

Solution

暴力枚举每10个连续字符,判断是否是合法日期并计数即可.

Code

var
  s,t:ansistring;
  n,i,j,d,m,y,k,max:longint;
  a:array[1..31,1..12,2013..2015]of longint;
  day:array[1..12] of longint;
begin
  readln(s);
  day[1]:=31;
  day[2]:=28;
  day[3]:=31;
  day[4]:=30;
  day[5]:=31;
  day[6]:=30;
  day[7]:=31;
  day[8]:=31;
  day[9]:=30;
  day[10]:=31;
  day[11]:=30;
  day[12]:=31;
  max:=-10;
  for i:=1 to length(s) do
  if (s[i]=‘-‘)and(s[i+3]=‘-‘)and((s[i-2]<=‘9‘)and(s[i-2]>=‘0‘))
  and((s[i-1]<=‘9‘)and(s[i-1]>=‘0‘))and((s[i+1]<=‘9‘)and(s[i+1]>=‘0‘))
  and((s[i+2]<=‘9‘)and(s[i+2]>=‘0‘))and((s[i+4]<=‘9‘)and(s[i+4]>=‘0‘))
  and((s[i+5]<=‘9‘)and(s[i+5]>=‘0‘))and((s[i+6]<=‘9‘)and(s[i+6]>=‘0‘))
  and((s[i+7]<=‘9‘)and(s[i+7]>=‘0‘))then
  begin
    d:=(ord(s[i-2])-48)*10+ord(s[i-1])-48;
    m:=(ord(s[i+1])-48)*10+ord(s[i+2])-48;
    y:=(ord(s[i+4])-48)*1000+(ord(s[i+5])-48)*100+(ord(s[i+6])-48)*10+ord(s[i+7])-48;
    if (y<=2015)and(y>=2013)and(m<=12)and(m>=1) then
    if (d>=1)and(d<=day[m]) then
    begin
      inc(a[d,m,y]);
    end;
  end;
  for i:=1 to 31 do
    for j:=1 to 12 do
      for k:=2013 to 2015 do
      if a[i,j,k]>max then
      begin
        max:=a[i,j,k];
        d:=i;m:=j;y:=k;
      end;
  if d<10 then write(‘0‘,d) else write(d);
   write(‘-‘);
  if m<10 then write(‘0‘,m) else write(m);
  write(‘-‘);
  writeln(y);
end.

C

Problem description

有n个箱子,进行一次操作,把一个箱子的球拿出来,然后放到后面的箱子里,每个箱子放一个,直到取的球放完. 若放到了第n个箱子,下一个放第1个箱子.给出操作后的情况和最后放球的箱子,求出原来的情况.

Data Limit：2?≤?n?≤?105  Time Limit: 1s

Solution

现在球数最少的的箱子,这就是原来选的箱子,然后就可以求出原来的样子啦.

Code

#include<cstdio>
long long n,i,j,x,min=1e18,ans=0;
long long a[500000];
int main()
{
    scanf("%lld%lld",&n,&x);
    for (i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        if (a[i]<min)min=a[i];
    }
    if (min>0)min--;
    for (i=1;i<=n;i++)a[i]=a[i]-min;
    while (a[x]>0)
    {
        ans++;a[x]--;x--;
        if (x==0)x=n;
    }
    a[x]=min*n+ans;
    for (i=1;i<=n;i++)printf("%lld ",a[i]);
    return 0;
}

D

Problem description

给出n个点的颜色(黑或白)和权值(所有与其相连的边的权值和)求出原来的树.

Data Limit：n <= 1e5  Time Limit: 1s

Solution

贪心,每次枚举两个点,连权值最大的边即可.

Code

#include<cstdio>
#include<algorithm>
using namespace std;
long long  n,i,j,t,topb,topw,ans[2000000][4],topa;
bool bo=false;
struct point
{
    long long a,sum;
};
bool bbb(point x,point y)
{
    return x.sum>y.sum;
}
point black[2000000],white[2000000];
int main()
{
    scanf("%lld",&n);
    for (i=1;i<=n;i++)
    {
        scanf("%lld",&t);
        if (t==0)
        {
            topw++;
            scanf("%lld",&white[topw].sum);
            white[topw].a=i;
        }else
        {
            topb++;
            scanf("%lld",&black[topb].sum);
            black[topb].a=i;
        }
    }
    i=1;j=1;topa=0;
    while ((i<=topb)&&(j<=topw))
    {
        topa++;
        ans[topa][1]=black[i].a;
        ans[topa][2]=white[j].a;

        t=min(white[j].sum,black[i].sum);
        ans[topa][3]=t;
        white[j].sum-=t;black[i].sum-=t;
        if (black[i].sum) j++;
        else if(white[j].sum) i++;
        else if(i<topb) i++;
        else j++;
    }
    for (i=1;i<=n-1;i++)
    printf("%lld %lld %lld\n",ans[i][1],ans[i][2],ans[i][3]);
    return 0;
}

E

Problem description

给出一个平面上的n个点,用两条平行于x轴的直线和两条平行于y轴的直线把平面分成9块 使每个块中的点符合给出的要求.

Data Limit：9?≤?n?≤?10^5 Time Limit:2s

Solution

先全排列出每个块的点数然后用线段树求出能否满足题意

Code

#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<queue>
#define inf 1000000005
#define M 40
#define N 100005
#define maxn 300005
#define eps 1e-12
#define zero(a) fabs(a)<eps
#define Min(a,b) ((a)<(b)?(a):(b))
#define Max(a,b) ((a)>(b)?(a):(b))
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define MOD 1000000007
#define lson step<<1
#define rson step<<1|1
#define sqr(a) ((a)*(a))
#define Key_value ch[ch[root][1]][0]
#define test puts("OK");
#define pi acos(-1.0)
#define lowbit(x) ((-(x))&(x))
#define HASH1 1331
#define HASH2 10001
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
struct Set_tree{
    int left,right;
    vector<int>v;
}L[N*4];
struct Point{
    int x,y;
    bool operator<(const Point n)const{
        return x!=n.x?x<n.x:y<n.y;
    }
}p[N];
int n,x[N],y[N];
int a[9],b[9];
double ret_x1,ret_x2,ret_y1,ret_y2;
void Bulid(int step,int l,int r){
    L[step].left=l;
    L[step].right=r;
    L[step].v.clear();
    for(int i=l;i<=r;i++)
        L[step].v.pb(p[i].y);
    sort(L[step].v.begin(),L[step].v.end());
    if(l==r)
        return;
    int m=(l+r)>>1;
    Bulid(lson,l,m);
    Bulid(rson,m+1,r);
}
int Query(int step,int l,int r,int val){
    if(L[step].left==l&&r==L[step].right){
        if(L[step].v.size()==0) return 0;
        if(L[step].v[0]>val) return 0;
        if(L[step].v.back()<=val) return L[step].v.size();
        return (upper_bound(L[step].v.begin(),L[step].v.end(),val)-L[step].v.begin());
    }
    int m=(L[step].left+L[step].right)>>1;
    if(r<=m) return Query(lson,l,r,val);
    else if(l>m) return Query(rson,l,r,val);
    else return Query(lson,l,m,val)+Query(rson,m+1,r,val);
}
bool ok(){
    int x1=b[a[0]]+b[a[1]]+b[a[2]]-1;
    int x2=x1+b[a[3]]+b[a[4]]+b[a[5]];
    int y1=b[a[0]]+b[a[3]]+b[a[6]]-1;
    int y2=y1+b[a[1]]+b[a[4]]+b[a[7]];
    if(x1+1>=n||x[x1]==x[x1+1]) return false;
    if(x2+1>=n||x[x2]==x[x2+1]) return false;
    if(y1+1>=n||y[y1]==y[y1+1]) return false;
    if(y2+1>=n||y[y2]==y[y2+1]) return false;
    if(Query(1,0,x1,y[y1])!=b[a[0]]) return false;
    if(Query(1,0,x1,y[y2])!=b[a[0]]+b[a[1]]) return false;
    if(Query(1,x1+1,x2,y[y1])!=b[a[3]]) return false;
    if(Query(1,x1+1,x2,y[y2])!=b[a[3]]+b[a[4]]) return false;
    ret_x1=(x[x1]+x[x1+1])/2.0;
    ret_x2=(x[x2]+x[x2+1])/2.0;
    ret_y1=(y[y1]+y[y1+1])/2.0;
    ret_y2=(y[y2]+y[y2+1])/2.0;
    return true;
}
int main(){
    //freopen("input.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d%d",&p[i].x,&p[i].y);
            x[i]=p[i].x;y[i]=p[i].y;
        }
        sort(p,p+n);
        sort(x,x+n);
        sort(y,y+n);
        Bulid(1,0,n-1);
        for(int i=0;i<9;i++) scanf("%d",&b[i]);
        for(int i=0;i<9;i++) a[i]=i;
        int t=362880;
        while(t--){
            if(ok()){
                printf("%.1f %.1f\n%.1f %.1f\n",ret_x1,ret_x2,ret_y1,ret_y2);
                break;
            }
            next_permutation(a,a+9);
        }
        if(t<=0) puts("-1");
    }
    return 0;
}