Codeforces round 171 div2

Http://codeforces.com/contest/279/problem/A

A: You can draw a drawing on your own to understand the law. Let's give you a point and ask how many turns this point has taken on the spiral.

Thought: I did it through simulation. Because the data volume is not large, it is violent from the inside out to check whether it is on the current line segment.

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;int dir[4][2]={1,0,0,1,-1,0,0,-1};int main(){    int x,y;    scanf("%d%d",&x,&y);    int sx=0,sy=0,ans=0,d=0,len=1,tx=0,ty=0;    while(1)    {        int num=2;        while(num--)        {            //printf("%d %d\n",tx,ty);            d%=4;            sx=tx,sy=ty;            tx=sx+len*dir[d][0],ty=sy+len*dir[d][1];            if(d%2)            {                if(x==sx&&(y-sy)*(y-ty)<=0)                {                    printf("%d\n",ans);                    return 0;                }            }            else            {                if(y==sy&&((x-sx)*(x-tx))<=0)                {                    printf("%d\n",ans);                    return 0;                }            }            ans++;            d++;        }        len++;    }    return 0;}

Http://codeforces.com/contest/279/problem/ B

B: Give You n numbers. The longest continuous interval is the sum of the numbers in the interval not greater than T.

Train of Thought: At first glance, I think it is better to have two points. We scan from left to right, set L to the left endpoint of the current range, initially 1, set the current traversal to I, then if sum [L, I]> T, then, I add L ++ until sum [L, I] <= T. At this time, I-l + 1 is the maximum value of the interval ending with I and is set to TMP, we can find the maximum value of TMP in the traversal process. The time complexity is O (n ).

#include <iostream>#include <string.h>#include <stdio.h>#define maxn 100010using namespace std;int a[maxn];int max(int a,int b){     return a>b?a:b;}int main(){    //freopen("dd.txt","r",stdin);    int n,t,i;    scanf("%d%d",&n,&t);    for(int i=n;i>=1;i--)    scanf("%d",&a[i]);    int l=0,sum=0,ans=0;    for(i=1;i<=n;i++)    {        sum+=a[i];        if(sum<=t)        ans=max(ans,i-l);        else        {            while(sum>t)            {                sum-=a[++l];            }         ans=max(ans,i-l);        }    }    printf("%d\n",ans);    return 0;}

Http://codeforces.com/contest/279/problem/C

C: give you n numbers a [1] ~ A [n]: ask if the range [L, R] is non-decreasing when the previous segment is met, and the latter segment is not increasing. (The first and second sections can be blank)

Idea: set MI [I] And ma [I] to indicate the left end of a [I], non-decreasing substring and the left end of non-incrementing substring respectively. This can be solved within the O (n) complexity. If Mi [R] <= L or ma [I] <= L, it indicates that the interval [L, R] is a non-decreasing string or a non-incrementing string, and the output is yes, otherwise, set Po = mi [R], indicating that the range [Po, R] is not an ascending string. If Ma [po] <= L, it indicates that the range [L, po] is not decreasing, [Po, R] is not incremental, and the output is yes; otherwise, no is output.

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#define maxn 100010using namespace std;int a[maxn];int ma[maxn],mi[maxn];int main(){    //freopen("dd.txt","r",stdin);    int n,i,q;    scanf("%d%d",&n,&q);    for(i=1;i<=n;i++)    scanf("%d",&a[i]);    ma[1]=mi[1]=1;    for(i=2;i<=n;i++)    {        if(a[i]>a[i-1])        {            ma[i]=ma[i-1];            mi[i]=i;        }        else if(a[i]<a[i-1])        {            ma[i]=i;            mi[i]=mi[i-1];        }        else        {            ma[i]=ma[i-1];            mi[i]=mi[i-1];        }    }    while(q--)    {        int l,r;        scanf("%d%d",&l,&r);        if(ma[r]<=l||mi[r]<=l)        {            printf("Yes\n");        }        else        {            int po=mi[r];            if(ma[po]<=l)            printf("Yes\n");            else            printf("No\n");        }    }    return 0;}

Http://codeforces.com/contest/279/problem/D

D: I read the conclusion report of becauseofyou. The Code basically references the report of becauseofyou. The first state is to compress DP.

The State is represented in binary format. 1 of the highest bits of State X is the I bit, indicating that a [I] is combined with the preceding number, 1 before the I bit indicates that the current State has saved a [J] (0 <= j <I and the J bit of X is 1 ), 0 indicates that the current status is not sure whether a [J] is saved (0 <= j <I and the J-bit of X is 0 ). set DP [x] to the minimum number of variables in the X state. In this case, we require DP [1 <(n-1)]. Because each number is obtained one by one, when the State X is obtained again, the previous State must contain the number of the previous requirement (that is, a [I-1]), if a [I] = A [J] + A [k] (0 <= j, k <I ), indicates that a [I] is transferred by a [J] And a [K], so the previous State must save a [J] And a [K], so the state transition equation came out.

Note that if num is set to State X and the number of 1 is expressed in binary, DP [x] should be at least num, which can be understood by the meaning of State X.

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#define inf 2100000000using namespace std;int a[25],pow[23];int dp[1<<23],n;void init(){    memset(dp,-1,sizeof(dp));    pow[0]=1;    for(int i=1;i<23;i++)    pow[i]=2*pow[i-1];}int min(int a,int b){    return a<b?a:b;}int max(int a,int b){    return a>b?a:b;}int dfs(int x){    if(dp[x]!=-1)    return dp[x];    int num=__builtin_popcount(x);    int ans=inf,tmp;    for(int i=n-1;i>=0;i--)    {        if(x&pow[i])        {            int tt=(x^pow[i])|pow[i-1];            for(int j=i-1;j>=0;j--)            {                for(int k=j;k>=0;k--)                {                    if(a[i]==a[j]+a[k])                    {                        tmp=dfs(tt|pow[j]|pow[k]);                        if(tmp!=inf)                        ans=min(ans,max(tmp,num));                    }                }            }            break;        }    }    dp[x]=ans;    return ans;}int main(){    //freopen("dd.txt","r",stdin);    int i;    init();    scanf("%d",&n);    for(i=0;i<n;i++)    scanf("%d",&a[i]);    dp[1]=1;    int ans=dfs(pow[n-1]);    if(ans==inf)    ans=-1;    printf("%d\n",ans);    return 0;}

Http://codeforces.com/contest/279/problem/E

E: Give You A binary representation of the number a [1] ~ A [n] is used to calculate the number in the form of least 2 ^ K and-2 ^ K. Their sum is equal to the given number.

Idea: greedy is also acceptable, but it is quite simple for me to do it through DP. A simple conclusion is that the K numbers used are different.

Set DP [I] to the minimum value required by the last I-digit. We traverse from the back to the front.

When a [I] = 1, we can directly add 2 ^ I, that is, DP [I] = DP [I + 1], or we can use 2 ^ (I + 1)-2 ^ J (0 <= j <I) to set the I bit to 1, but this is not complete yet, we also need to add DP [n-J + 1], and for 0 in [I, n-J, we also use-2 ^ X (I <x <= N-J and a [x] = 0) to minimize the number obtained, it is necessary to make the DP [n-J + 1] + 2 + num (Num is the number of 0 in the range [I, n-J]) the smallest, we set this value to W [x] for different positions, which seems to require O (N ^ 2) Complexity. Otherwise, we can find that, if the position Po is the smallest, in fact, if a [I] = 1, num remains the same, W [po] is the smallest.

If a [I] = 0, num ++, W [po] ++. At this time, only w [I] may be less than W [po]. Let's compare the update. The final complexity can reach O (n ).

(The expression capability is the same. I don't understand it at all. Let's look at the code)

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;char str[1000010];int dp[1000010];int min(int a,int b)  {    return a>b?b:a;    }int main(){   // freopen("dd.txt","r",stdin);  int i,tmp=0;  scanf("%s",str+1);  int n=strlen(str+1);  dp[n]=str[n]-'0';  for(i=n-1;i>=1;i--)    {      if(str[i]=='1')        dp[i]=min(dp[i+1]+1,tmp+2);      else        {          tmp++;          dp[i]=dp[i+1];          tmp=min(tmp,dp[i]);          }      }  printf("%d\n",dp[1]);  return 0;}