5A,5B直接实现，没有什么难度，主要是要读懂题，例如5B...
5C dp可惜我又没想出来，好伤感...菜鸡的春天在哪里啊...
看了下官方的tutorial，讲的很清楚，两个数组d[n],c[n]
d[i]，记录i位置‘）’对应的‘（’位置，当然前提是要第i个是‘（’，如果不存在则d[i]=-1;
c[i]，记录以i处的‘）’为结尾的最长子串的第一个‘（’的位置，如果不存在则-1;
具体实现要维护一个栈，从0开始，如果是‘(,入栈所在位置;如果是')',使得c[i]=d[i]=stack.top();如果stack为空c[i]=d[i]=-1;
同时，判断d[i]-1处是不是')'，如果是，那么判断c[d[i]-1]是不是-1,不是的话,c[i]=c[d[i]-1];
大概这样子,最后i-c[i]+1的最大值即为最大,记得-1处要舍去啊...
我解释的不够清楚,就把tutorial 也贴出来
First of all, for each closing bracket in our string let's define 2 values:
- d[j] = position of corresponding open bracket, or -1 if closing bracket doesn't belong to any regular bracket sequence.
- c[j] = position of earliest opening bracket, such that substring s(c[j], j) (both boundaries are inclusive) is a regular bracket sequence. Let's consider c[j] to be -1 if closing bracket doesn't belong to any regular bracket sequence.
It can be seen, that c[j] defines the beginning position of the longest regular bracket sequence, which will end in position j. So, having c[j] answer for the problem can be easily calculated.
Both d[j] and c[j] can be found with following algorithm, which uses stack.
- Iterate through the characters of the string.
- If current character is opening bracket, put its position into the stack.
- If current character is closing bracket, there are 2 subcases:
- Stack is empty - this means that current closing bracket doesn't have corresponding open one. Hence, both d[j] and c[j] are equal to -1.
- Stack is not empty - we will have position of the corresponding open bracket on the top of the stack - let's put it to d[j] and remove this position from the stack. Now it is obvious, that c[j] is equal at least to d[j]. But probably, there is a better value for c[j]. To find this out, we just need to look at the position d[j] - 1. If there is a closing bracket at this position, and c[d[j] - 1] is not -1, than we have 2 regular bracket sequences s(c[d[j] - 1], d[j] - 1) and s(d[j], j), which can be concatenated into one larger regular bracket sequence. So we put c[j] to be c[d[j] - 1] for this case.
5D 纯数学问题,满满算,细心点.
5A #include<iostream> #include<list> #include<string> using namespace std; int main() { list<string>strList; int traffic=0; string str; while(getline(cin,str)) { if(str[0]=='+') strList.push_back(str.substr(1)); if(str[0]=='-') { for(list<string>::iterator it=strList.begin();it!=strList.end();it++) { if(str.substr(1)==*it) { strList.erase(it); break; } } } if(str[0]!='+'&&str[0]!='-') { int i=str.find(':'); string msg=str.substr(i+1); traffic+=strList.size()*msg.length(); } } cout<<traffic<<endl; return 0; }
5B #include<iostream> #include<string> #include<vector> using namespace std; int main() { vector<string>vs; string str; int maxlen=0; while(getline(cin,str)) {A if(str.length()>maxlen) maxlen=str.length(); vs.push_back(str); } for(int i=0;i<maxlen+2;i++) { cout<<'*'; } cout<<endl; int m=0; for(int i=0;i<vs.size();i++) { int l,r,s; s=maxlen-vs[i].length(); if(s%2) m++; if(m%2==0) { l=s%2?s/2+1:s/2; r=s/2; } else { r=s%2?s/2+1:s/2; l=s/2; } cout<<'*'; for(int j=0;j<l;j++) cout<<' '; cout<<vs[i]; for(int j=0;j<r;j++) cout<<' '; cout<<'*'; cout<<endl; } for(int i=0;i<maxlen+2;i++) { cout<<'*'; } cout<<endl; return 0; }
5C #include<iostream> #include<fstream> #include<stack> #include<string> using namespace std; int main() { stack<int> sta; int *c=new int[1000002](); int *d=new int[1000002](); string str; cin>>str; for(int i=0;i<str.length();i++) { if(str[i]=='(') sta.push(i); else { if(sta.empty()) { c[i]=-1; d[i]=-1; } else { d[i]=sta.top(); c[i]=sta.top(); sta.pop(); if(d[i]-1>=0&&str[d[i]-1]==')'&&c[d[i]-1]!=-1) { c[i]=c[d[i]-1]; } } } } int maxlen=0,cnt=0; for(int i=0;i<str.length();i++) { if(c[i]==-1||str[i]=='(') continue; if(i-c[i]+1>maxlen) { maxlen=i-c[i]+1; cnt=1; } else { if(i-c[i]+1==maxlen) cnt++; } } if(maxlen==0) cout<<0<<" "<<1<<endl; else cout<<maxlen<<" "<<cnt<<endl; return 0; } //int main() //{ // int maxlen=0,start=-1,maxcnt=0; // stack<char>sc; // ifstream fin("testdata.txt"); // string str; // fin>>str; // for(int i=0;i<str.length();) // { // if(str[i]=='(') // { // int left=1,right=0; // int j=i+1; // int nowmax=0; // int index=0; // while(j<str.length()) // { // if(str[j]=='(') // { // left++; // } // else // { // right++; // } // j++; // if(right>left) // break; // if(left>right) // { // if(right==1) // { // left--; // right=0; // nowmax+=2; // index=j-1; // } // continue; // } // if(left==right) // { // nowmax+=2; // left=0; // right=0; // index=j-1; // } // } // if(nowmax>maxlen) // { // maxlen=nowmax; // maxcnt=1; // } // else // { // if(nowmax==maxlen) // { // maxcnt++; // } // } // i=i+(nowmax>1?index+1:1); // // } // else // { // i++; // } // } // // if(maxlen==0) // { // cout<<maxlen<<" "<<1<<endl; // } // else // { // cout<<maxlen<<" "<<maxcnt<<endl; // } // //}
5D #include<iostream> #include<cmath> #include<iomanip> using namespace std; int main() { int v,a,l,d,w; cin>>a>>v>>l>>d>>w; if(w>=v) { double t=(double)v/a; if(t*t*a/2>=l) { cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<sqrt(((double)l*2)/a)<<endl; } else { double left=l-t*t*a/2; cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<left/v+t<<endl; } } else { double t=(double)v/a; double t1=(double)w/a; if(t1*t1*a/2>=d) { if(t*t*a/2>=l) cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<sqrt(((double)l*2)/a)<<endl; else { double left=l-t*t*a/2; cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<left/v+t<<endl; } } else { double dleft=d-t1*t1*a/2; double dt=(double)(v-w)/a; double dmid=(w+v)*dt/2; if(dleft/2<=dmid) { double t2=sqrt((dleft+(double)w*w/a)/a)-(double)w/a; double acet=(double)(v-w)/a; if((v+w)*acet/2>=l-d) { double t3=sqrt((2*l-2*d+(double)w*w/a)/a)-(double)w/a; cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<t1+2*t2+t3<<endl; } else { double t3=(double)(v-w)/a; double t4=(l-d-(v+w)*t3/2)/v; cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<t1+2*t2+t3+t4<<endl; } } else { double t2=(double)(v-w)/a*2+(dleft-2*dmid)/v; double acet=(double)(v-w)/a; if((v+w)*acet/2>=l-d) { double t3=sqrt((2*l-2*d+(double)w*w/a)/a)-(double)w/a; cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<t1+t2+t3<<endl; } else { double t3=(double)(v-w)/a; double t4=(l-d-(v+w)*t3/2)/v; cout<<setiosflags(ios_base::fixed)<<setprecision(5)<<t1+t2+t3+t4<<endl; } } } } return 0; }