之前挂上的
今天填坑
2018.2.14 #462
A
给两个集合，B分别可以从一个集合中选一个数，B想乘积最大，A想最小，A可以删除一个第一个集合中的元素，问最小能达到多少。
这题。。水死啦。我居然还wa了一发，因为ans初值定的0.。。。。我真是傻。
n3暴力就行
#include<bits/stdc++.h> using namespace std; #define ll long long #define inf 1000000000000000001ll int n,m; ll a[100],b[100]; ll ans=inf; int main(){ cin>>n>>m; for(int i=1;i<=n;i++)cin>>a[i]; for(int i=1;i<=m;i++)cin>>b[i]; for(int k=1;k<=n;k++){ ll ret=-inf; for(int i=1;i<=n;i++){ if(i==k)continue; for(int j=1;j<=m;j++){ ret=max(ret,a[i]*b[j]); } } ans=min(ans,ret); } cout<<ans<<endl; return 0; }
B
我记得小时候的脑筋急转弯
6666=4
8888=8
1234=1
5678=?
哈哈
数圈嘛
先全输出8，剩一个就输出6，剩下的输出1
#include<bits/stdc++.h> using namespace std; #define ll long long int k; int main(){ cin>>k; if(k>36){ cout<<-1<<endl; return 0; } else { for(int i=1;i<=18;i++){ if(k>=2){ cout<<8; k-=2; } else if(k==1){ cout<<6;k--;} else cout<<1; } } return 0; }
C
一个只有1，2的序列，你可以翻转一次任意子串，问翻转之后最长不下降子序列。
选择很多？？？
这题随便做嘛。
我是利用了一个小性质，就是从1到2的分界点，一定是在翻转的区间里，否则没意义。
然后枚举翻转区间，维护一下最长不下降子序列就好了。
#include<bits/stdc++.h> using namespace std; #define ll long long int a[2010]; int l1[2010],l2[2010],r1[2010],r2[2010]; int n; int ans; int main(){ cin>>n; for(int i=1;i<=n;i++)cin>>a[i]; for(int i=1;i<=n;i++){ l1[i]=l1[i-1]; l2[i]=l2[i-1]; if(a[i]==1)l1[i]++; else l2[i]++; } for(int i=n;i>=1;i--){ r1[i]=r1[i+1]; r2[i]=r2[i+1]; if(a[i]==1)r1[i]++; else r2[i]++; } for(int i=1;i<=n+1;i++){ ans=max(ans,l1[i-1]+r2[i]); int ret=l1[i-1]+r2[i]; int j=i-1; while(j&&a[j]==1)j--; int tt=0; for(;j>0;j--){ if(a[j]==2)tt++; else if(tt)tt--; ans=max(ans,ret+tt); } } cout<<ans<<endl; return 0; }
D
好题！！！
n order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.
Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...
Given two integers p and k, find a polynomial f(x) with non-negative integer coefficients strictly less than k, whose remainder is p when divided by (x + k). That is, f(x) = q(x)·(x + k) + p, where q(x) is a polynomial (not necessarily with integer coefficients).
The only line of input contains two space-separated integers p and k (1 ≤ p ≤ 1018, 2 ≤ k ≤ 2 000).
If the polynomial does not exist, print a single integer -1, or output two lines otherwise.
In the first line print a non-negative integer d — the number of coefficients in the polynomial.
In the second line print d space-separated integers a0, a1, ..., ad - 1, describing a polynomial 
fulfilling the given requirements. Your output should satisfy 0 ≤ ai < k for all 0 ≤ i ≤ d - 1, and ad - 1 ≠ 0.
If there are many possible solutions, print any of them.
这题没有special judge
只有唯一解
根据多项式意义，可以发现，f(x)的第一位是可以立即确定的，然后根据第一位可以推出第二位....依次类推。
然后我们发现，这个不断从上一项推出下一项的过程，很像之前做过的负进制转换。（luoguP 1107）
吐槽一下，我当时发现很像，并且过了pretest之后，告诉洛谷群，结果群里面一堆小孩说我钓鱼。。。。。
这件事情教育我们，干什么要先过脑子，，，，，
其实就是把p转化为（-k）进制，输出，结束。
(自行去luogu题解粘代码)
E
计算几何，，，，，gg
#include<bits/stdc++.h> using namespace std; struct point{ double x,y; }; double xmult(point p1,point p2,point p0){ return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } double dist(point p1,point p2){ return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } point intersection(point u1,point u2,point v1,point v2){ point ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; return ret; } const double eps=1e-8; point dot_to_circle(point c,double r,point p){ point u,v; if (dist(p,c)<eps) return p; u.x=c.x+r*fabs(c.x-p.x)/dist(c,p); u.y=c.y+r*fabs(c.y-p.y)/dist(c,p)*((c.x-p.x)*(c.y-p.y)<0?-1:1); v.x=c.x-r*fabs(c.x-p.x)/dist(c,p); v.y=c.y-r*fabs(c.y-p.y)/dist(c,p)*((c.x-p.x)*(c.y-p.y)<0?-1:1); return dist(u,p)<dist(v,p)?u:v; } void intersection_line_circle(point c,double r,point l1,point l2,point& p1,point& p2){ point p=c; double t; p.x+=l1.y-l2.y; p.y+=l2.x-l1.x; p=intersection(p,c,l1,l2); t=sqrt(r*r-dist(p,c)*dist(p,c))/dist(l1,l2); p1.x=p.x+(l2.x-l1.x)*t; p1.y=p.y+(l2.y-l1.y)*t; p2.x=p.x-(l2.x-l1.x)*t; p2.y=p.y-(l2.y-l1.y)*t; } void intersection_circle_circle(point c1,double r1,point c2,double r2,point& p1,point& p2){ point u,v; double t; t=(1+(r1*r1-r2*r2)/dist(c1,c2)/dist(c1,c2))/2; u.x=c1.x+(c2.x-c1.x)*t; u.y=c1.y+(c2.y-c1.y)*t; v.x=u.x+c1.y-c2.y; v.y=u.y-c1.x+c2.x; intersection_line_circle(c1,r1,u,v,p1,p2); } int n; int sqr(int x){ return x*x; } int cross(int a,int b,int c,int d,int e,int f){ if(sqr(a-d)+sqr(b-e)>=sqr(c+f))return 0; if(sqr(a-d)+sqr(b-e)<=sqr(c-f))return 0; return 1; } int a,b,c,d,e,f,g,h,i; int ct(int a,int b,int c,int d,int e,int f){ if(sqr(a-d)+sqr(b-e)==sqr(c+f))return 1; if(sqr(a-d)+sqr(b-e)==sqr(c-f))return 2; return 0; } //3 //-4 4 4 //2 4 2 //-1 0 6 int main() { cin>>n; if(n==1){ cin>>a>>b>>c; cout<<2; return 0; } if(n==2){ cin>>a>>b>>c>>d>>e>>f; cout<<3+cross(a,b,c,d,e,f); return 0; } cin>>a>>b>>c>>d>>e>>f>>g>>h>>i; int ret=4; if(cross(a,b,c,d,e,f)&&cross(d,e,f,g,h,i)&&cross(g,h,i,a,b,c)){ ret=8; double num1,num2,num3,num4,num5,num6; double r1,r2,r3; num1=a,num2=b,r1=c; num3=d,num4=e,r2=f; num5=g,num6=h,r3=i; point w,ww,p1,p2; w.x=a,w.y=b;ww.x=d,ww.y=e; intersection_circle_circle(w,r1,ww,r2,p1,p2); num1=p1.x,num2=p1.y,num3=p2.x,num4=p2.y; if(fabs((num1-num5)*(num1-num5)+(num2-num6)*(num2-num6)-r3*r3)<eps)ret--; if(fabs((num3-num5)*(num3-num5)+(num4-num6)*(num4-num6)-r3*r3)<eps)ret--; cout<<ret<<endl; } else { ret=4+cross(a,b,c,d,e,f)+cross(d,e,f,g,h,i)+cross(g,h,i,a,b,c); if(ret==6){ if((ct(a,b,c,d,e,f)+1)/2+(ct(d,e,f,g,h,i)+1)/2+(ct(g,h,i,a,b,c)+1)/2){ double num1,num2,num3,num4,num5,num6; double r1,r2,r3; num1=a,num2=b,r1=c; num3=d,num4=e,r2=f; if(cross(a,b,c,d,e,f)){ point w,ww,p1,p2; w.x=a,w.y=b;ww.x=d,ww.y=e; intersection_circle_circle(w,r1,ww,r2,p1,p2); num1=p1.x,num2=p1.y,num3=p2.x,num4=p2.y; num5=g,num6=h,r3=i;if(fabs((num1-num5)*(num1-num5)+(num2-num6)*(num2-num6)-r3*r3)>eps&&fabs((num3-num5)*(num3-num5)+(num4-num6)*(num4-num6)-r3*r3)>eps)ret++; } else if(cross(a,b,c,g,h,i)){point w,ww,p1,p2; w.x=a,w.y=b;ww.x=g,ww.y=h; intersection_circle_circle(w,r1,ww,i,p1,p2); num1=p1.x,num2=p1.y,num3=p2.x,num4=p2.y; num5=d,num6=e,r3=f;if(fabs((num1-num5)*(num1-num5)+(num2-num6)*(num2-num6)-r3*r3)>eps&&fabs((num3-num5)*(num3-num5)+(num4-num6)*(num4-num6)-r3*r3)>eps)ret++; } } } else if(ret==5){ ret+=((ct(a,b,c,d,e,f)+1)/2+(ct(d,e,f,g,h,i)+1)/2+(ct(g,h,i,a,b,c)+1)/2)/2; } if(ret==4){ if(ct(a,b,c,d,e,f)&&ct(d,e,f,g,h,i)&&ct(g,h,i,a,b,c)){ double num1,num2,num3,num4; if(ct(a,b,c,d,e,f)==1)num1=a+1.0*(d-a)*c/(c+f),num2=b+1.0*(e-b)*c/(c+f); else { if(c>f){ num1=a+1.0*(d-a)*c/(c-f),num2=b+1.0*(e-b)*c/(c-f); } else { num1=d+1.0*(a-d)*f/(f-c),num2=e+1.0*(b-e)*f/(f-c); } } if(ct(g,h,i,a,b,c)==1)num3=a+1.0*(g-a)*c/(c+i),num4=b+1.0*(h-b)*c/(c+i); else { if(c>i){ num3=a+1.0*(g-a)*c/(c-i),num4=b+1.0*(h-b)*c/(c-i); } else { num3=d+1.0*(a-g)*i/(i-c),num4=e+1.0*(b-h)*i/(i-c); } } if(fabs(num3-num1)>eps||fabs(num4-num2)>eps)ret++; } } cout<<ret<<endl; } return 0; }
为什么老是说我颓废呢，，，，，我不颓废啊。
感觉，带修改莫队真是简单。
codeforces round#466
因为F一直不会，当时想到莫队，怎奈不会修改。
其实很简单，中午用了3分钟就看明白待修改莫队了。
然后F题就变成了大水题。。。
这场比赛应该AK的，，，，而我却没把握住机会。
多少有点遗憾吧。
A
给一个大小为n的集合，和一个常数d，问最少删除多少个数才能使得集合中，最大值与最小值相差小于k。
(不懂Doggu辣么强的选手为什么没有秒切并1A)
cf的div2T1从来不用什么算法。。。
sort之后枚举最大最小值即可。
#include<bits/stdc++.h> using namespace std; int a[110]; int n,d; int ans=10000; int main(){ cin>>n>>d; for(int i=1;i<=n;i++){ cin>>a[i]; } sort(a+1,a+n+1); for(int i=1;i<=n;i++){ int ret=i-1; for(int j=n;j>i;j--){ if(a[j]-a[i]<=d)break; ret++; } ans=min(ans,ret); } cout<<ans; return 0; }
B
给4个整数，n,k,A,B。
n每次可以-1或/k，代价分别是A,B问用最小代价使n变成1
不说了，直接上代码。都能看懂。
#include<bits/stdc++.h> using namespace std; #define ll long long ll n,k,a,b; ll ans=0; int main(){ cin>>n>>k>>a>>b; if(k==1){ ans=(n-1)*a; cout<<ans<<endl; return 0; } while(n!=1){ if(n%k==0){ ll tt=(n-n/k)*a; ll t=b; ans+=min(t,tt); n=n/k; } else { if(n>k){ ans+=(n%k)*a; n-=n%k; } else { ans+=(n-1)*a; n=1; } } } cout<<ans; return 0; }
(Doggu这次没打好存粹是心态问题)
C
太水了，不说了，不说了。
前三题绝对是普及组难度的。
D
"We've tried solitary confinement, waterboarding and listening to Just In Beaver, to no avail. We need something extreme."
"Little Alena got an array as a birthday present..."
The array b of length n is obtained from the array a of length n and two integers l and r (l ≤ r) using the following procedure:
b1 = b2 = b3 = b4 = 0.
For all 5 ≤ i ≤ n:
- bi = 0 if ai, ai - 1, ai - 2, ai - 3, ai - 4 > r and bi - 1 = bi - 2 = bi - 3 = bi - 4 = 1
- bi = 1 if ai, ai - 1, ai - 2, ai - 3, ai - 4 < l and bi - 1 = bi - 2 = bi - 3 = bi - 4 = 0
- bi = bi - 1 otherwise
You are given arrays a and b' of the same length. Find two integers l and r (l ≤ r), such that applying the algorithm described above will yield an array b equal to b'.
It's guaranteed that the answer exists.
维护四个数，maxl,maxr,minl,minr。不断缩小范围即可。
#include<bits/stdc++.h> using namespace std; #define ll long long #define inf 1000000000 int n; int a[100010]; string b; int maxl,maxr,minl,minr; int main(){ maxl=maxr=inf; minl=minr=-inf; cin>>n; for(int i=0;i<n;i++)cin>>a[i]; cin>>b; bool f1,f2; int L,R; for(int i=4;i<n;i++){ f1=f2=0; if(b[i-1]=='0'&&b[i-2]=='0'&&b[i-3]=='0'&&b[i-4]=='0')f1=1; if(b[i-1]=='1'&&b[i-2]=='1'&&b[i-3]=='1'&&b[i-4]=='1')f2=1; L=min(min(a[i],min(a[i-1],a[i-2])),min(a[i-3],a[i-4])),R=max(max(a[i],max(a[i-1],a[i-2])),max(a[i-3],a[i-4])); int x=b[i]-'0'; int y=b[i-1]-'0'; if(x==1){ if(y==1){ if(f2){ minr=max(L,minr); } } else { if(f1){ minl=max(R+1,minl); } } } else { if(y==1){ if(f2){ maxr=min(maxr,L-1); } } else { if(f1){ maxl=min(maxl,R); } } } } cout<<minl<<" "<<maxr<<endl; return 0; }
E
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the 
smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
很容易得到一个小性质，选长度恰好为c的区间最优。
所以，st表求区间最小值。
dp[i]表示到i位置最大能刨去多大的数
dp[i]=max(dp[i-1],dp[i-10]+qmin(i-10+1,i))
ans=SUM-dp[n];
#include<bits/stdc++.h> using namespace std; #define ll long long int rmq[20][100010]; int lg[100010]; int n,c; ll dp[100010]; ll sum; int query(int l,int r){ int k=lg[r-l+1]; return min(rmq[k][l],rmq[k][r-(1<<k)+1]); } int main(){ cin>>n>>c; for(int i=2;i<=n;i++)lg[i]=lg[i>>1]+1; for(int i=1;i<=n;i++){ cin>>rmq[0][i]; sum+=rmq[0][i]; } for(int i=1;i<=18;i++){ for(int j=1;j+(1<<i)-1<=n;j++){ rmq[i][j]=min(rmq[i-1][j],rmq[i-1][j+(1<<i-1)]); } } for(int i=c;i<=n;i++){ dp[i]=max(dp[i-1],dp[i-c]+query(i-c+1,i)); } cout<<sum-dp[n]; return 0; }
F
带修改莫队裸题。
小结论，ans不超过2*sqrt(n)
#include<bits/stdc++.h> using namespace std; #define N 400010 map<int,int> ma; inline int read(){ int x=0;char ch=getchar(); while(ch<'0'||ch>'9')ch=getchar(); while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar(); return x; } int dd; struct Q{ int l,r,t,id; }q[N]; struct C{ int pos,x,o; }c[N]; int block; bool comp(Q u,Q v){ return u.l/block<v.l/block || u.l/block==v.l/block && u.r/block<v.r/block || u.l/block==v.l/block && u.r/block==v.r/block && u.t<v.t; } int a[N]; int b[N]; int n,m; int d[N]; int e[N]; void add(int x,int y){ b[e[x]]--; e[x]+=y; b[e[x]]++; } int ans[N]; int main(){ n=read(),m=read(); for(int i=1;i<=n;i++){ a[i]=read(); if(ma.count(a[i]))a[i]=ma[a[i]]; else ma[a[i]]=++dd,a[i]=dd; d[i]=a[i]; } block=ceil(pow(n,2./3)); int cnt=0; int cnt1=0; int x,y,z; for(int i=1;i<=m;i++){ x=read(),y=read(),z=read(); if(x==1)q[++cnt].l=y,q[cnt].r=z,q[cnt].t=cnt1,q[cnt].id=cnt; else { if(ma.count(z))z=ma[z]; else ma[z]=++dd,z=dd; c[++cnt1].pos=y,c[cnt1].x=z,c[cnt1].o=d[y];d[y]=z; } } int l,r,t; l=r=t=0; sort(q+1,q+cnt+1,comp); for(int i=1;i<=cnt;i++){ while(r<q[i].r){ r++; add(a[r],1); } while(l>q[i].l){ l--; add(a[l],1); } while(r>q[i].r){ if(r<=n)add(a[r],-1); r--; } while(l<q[i].l){ if(l)add(a[l],-1); l++; } while(t<q[i].t){ ++t; int pos=c[t].pos; a[pos]=c[t].x; if(pos>=l&&pos<=r)add(c[t].o,-1),add(c[t].x,1); } while(t>q[i].t){ int pos=c[t].pos; a[pos]=c[t].o; if(pos>=l&&pos<=r)add(c[t].o,1),add(c[t].x,-1); t--; } for(int j=1;j<=1000;j++){ if(!b[j]){ ans[q[i].id]=j; break; } } } for(int i=1;i<=cnt;i++){ printf("%d\n",ans[i]); } return 0; }
