2118: 卡片
时间限制: 1 Sec 内存限制: 128 MB题目描述
lrb喜欢玩卡牌。他手上现在有张牌，每张牌的颜色为红绿蓝中的一种。现在他有两种操作。一是可以将两张任意位置的不同色的牌换成一张第三种颜色的牌；二是可以将任意位置的两张相同颜色的牌换成一张该颜色的牌。两个操作后都可以将生成的牌放到任意位置。现在他想知道，最后一张牌可能是什么颜色的。
输入
第一行输入一个n，表示卡牌数量。
第二行输入一个由’B’,’G’,’R’组成的长度为n的字符串，分别表示卡牌的颜色为蓝色、绿色、红色中的一种。
输出
输出’B’,’G’,’R’中的若干个字母，按字典序输出。代表可能的最后一张牌的颜色。
样例输入
3 GRG
样例输出
BR
well,this problem was easy,but I made the order of the letters wrong.
so...you know...i fell to pass it.
the thinking doesn't need any explication.
here is the code
#include<iostream> using namespace std; string w; char a[3]={'B','G','R'};int a1[3]={0}; int main(){ int n;cin>>n;cin>>w; for(int i=0;i<n;i++)for(int j=0;j<3;j++)if(w[i]==a[j])a1[j]++; for(int i=0;i<3;i++)if(a1[i]==n){cout<<a[i];return 0;} if(n==2)for(int i=0;i<3;i++)if(!a1[i]){cout<<a[i];return 0;} for(int i=0;i<3;i++) if(a1[i]==n-1){for(int j=0;j<3;j++)if(j!=i)cout<<a[j];return 0;} for(int i=0;i<3;i++)cout<<a[i];return 0; }
the only problem is that when I used getchar optimization,it went wrong.
maybe this web never allows it...
2119: 取数
时间限制: 1 Sec 内存限制: 128 MB题目描述
lrb目前有n个数字，他想知道这n个数中选出若干个数，平均数减中位数的最大值是多少。
输入
第一行为n，为数字个数。
第二行有n个数，分别代表n个数字。
输出
输出一个数，为平均数减中位数的最大值，保留两位小数。
样例输入
4 1 2 3 4
样例输出
0.33
提示
对于100%的数据，n<=10^5,0<=a[i]<=10^6
well the code is the best language
#include <cstdio> #include <algorithm> #define cint register int #define fr(i,a,b) for(cint i=a;i<=b;i++) #define INF 1e9 #define ll long long //---------------read optimization----------------------------------- #define getchar() (SS==TT&&(TT=(SS=BB)+fread(BB,1,1<<15,stdin),SS==TT)?EOF:*SS++) char BB[1<<15],*SS=BB,*TT=BB; int read(){cint x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while('0'<=ch&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}return x*f;} //---------------optimization above----------------------------------- using namespace std; int n,a[200010]; ll maxw,maxl,left,right,w1,w2,p[200010]; double v1,v2,v,maxh=-1; double solute(ll w,ll l){return (double)(p[w]-p[w-l-1]+p[n]-p[n-l])/(double)(2*l+1)-(double)a[w];} //calculate out the average_number subtract median int TPM(int i){left=0;right=min(i-1,n-i); while(left+1<=right){//if has not run out the section w1=(right-left)/3+left;w2=right-(right-left)/3; v1=solute(i,w1);v2=solute(i,w2); if(v1>v2)right=w2-1;else left=w1+1;} }//Three_Point_Method void pans(double key){printf("%.2lf ",key);}//out the data int main(){n=read(); // read the total_sum of the array fr(i,1,n)a[i]=read();//read the array sort(a+1,a+n+1);//set the array in order fr(i,1,n)p[i]=p[i-1]+a[i];//mark each prefix sum if(n<=2){pans(0);return 0;} //the special case fr(i,1,n){TPM(i);v=solute(i,left);//mark the solution of the_average_number_subtract_median if(v>maxh)maxh=v,maxw=i,maxl=left;}//if the solution is the best ,renew it. pans(maxh);return 0;//out the ans }
2120: 密码
时间限制: 1 Sec 内存限制: 128 MB题目描述
lrb去柜员机取钱，并输入了他的四位密码。但是这一切都被一个黑帮拍摄了下来。幸好，这一次他的银行卡里并没有剩余任何钱。lrb知道了他的账户被异常登录后十分害怕，然后修改了他的密码。从此以后他为了不被看出密码，他开始“徐晃”。但这一切还是被拍摄了下来，拍摄多次以后，这个黑帮找到了你，希望你在一秒内帮他算出lrb的可能使用的密码数量。
输入
第一行输入一个n，为lrb被拍摄的动作次数。
接下来n行，每行第一个数为t，表示lrb接下来的手指移动次数，接下来为一个长度为t的数字串，表示lrb手指经过的轨迹。lrb手指经过的位置，可能没有按，也有可能按了多次。
输出
输出一行，为可能的密码数量。
样例输入
2 3 123 3 234
样例输出
5
提示
lrb可能用的是2222,2223,2233,2333,3333五种密码
设所有轨迹串的总长度为L。
对于100%的数据，1<=n<=1000,1<=t<=10^4,L<=10^6
the code is the best language...
#include <bits/stdc++.h> #define ll long long using namespace std; #define cint register int #define cchar register char #define fr(cint,i,a,b) for(cint i=a;i<=b;i++) #define fd(i,a,b) for(cint i=a;i>=b;i--) //---------------read optimization----------------------------------- char BB[1<<15],*SS=BB,*TT=BB; short inline read(){short x=0;char ch=getchar(); while(ch<'0'||ch>'9')ch=getchar(); while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}return x;} //---------------optimization above----------------------------------- int n,cnt; //n:just the total_num //cnt:mark the final solution(just the possible number of password) short k[1005],ans[5]={0,1,3,3,1}; //k:use to read the move frequency char *ch[1005],a[5]; //a:use to mark the password //ch:use to read the given trail struct sth{short las[10];void init(){memset(las,0,sizeof(las));} }*suf[1004],tmp; //las's 1 to 9 individually mark the last_appear_site of each number bool check(int x){int w,fr;//set two variable fr(cint,i,1,n){//make a loop from 1 to n for(w=1,fr=0;w<=x;w++)//initializate fr as 0 //and give w a loop from 1 to x if(suf[i][fr].las[a[w]])fr=suf[i][fr].las[a[w]]; //if the last_appear_site is real,just renew the fr else break; //or just means that the site isn't real, //just means that this suppose is wrong just this case can't be real //just break if(w<=x)return false;//NO SOLUTION case,just return false }return true;//or,this case is real ,just return true } void dfs(int x){ fr(cchar,i,0,9)//make a loop from 0 to 9(type:char) if(i!=a[x-1]){//if the number has not been marked a[x]=i;//just mark it and then go on if(check(x))cnt+=ans[x];//if after check this case is true ,just add the ans if(x<4)dfs(x+1);//because the array:ans's section is from 0 to 4 //so we can only operate when it was inside the section, //if x<4,just means that x+1<=4,means satisfy the requirement //if satisfy this,just go on to the next operatation }} //deep search int main(){n=read();//read the total_num of the trail fr(cint,i,1,n)//make a loop from 1 to n { k[i]=read();//read the move frequency ch[i]=new char[k[i]+2];suf[i]=new sth[k[i]+2];//set the memory dynamically //one point is very important,the new_type[size]_operation only support pointer scanf("%s",ch[i]+1);tmp.init();//read the trail and initialilzate the letters' array fd(j,k[i],0){//make a loop from k[i] down to 0 suf[i][j]=tmp;//well,set the pointer:suf against tmp //this operation just means that send the tmp's address to suf[i][j] //afer this operation,you can use suf[i][j].lase[] to replace the tmp.las[] if(j)//if j is ont zero just means that it is inside the section //just go on tmp.las[ch[i][j]-'0']=j;//renew the last_appear_site } } a[0]=-1;dfs(1);//initializate the a[0] as a boundary,and then go to the operate_function //!-_-!,the highlight part!!!! printf("%d",cnt);return 0;//out the ans }