codejam 2016 Round A APAC Test

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1.Googol String 对半去找点，用flag标记一下对称的次数。

int main(){    int Tcase;    if(freopen("A-large-practice.in","r",stdin)==NULL)printf("failed!\n");    if(freopen("A-large-practice.out","w",stdout)==NULL)printf("failed!\n");    long long k;    long long len[100];    len[0] = 0;    int lenid = 0;    while(len[lenid] < 1e18){        len[lenid+1] = len[lenid] * 2 + 1;        lenid++;    }    long long ans;    bool flag;    int tid;    while(scanf("%d",&Tcase) == 1){        for(int tcase = 1;tcase<= Tcase;tcase++){            scanf("%lld",&k);            if(k == 1) ans = 0;            else{                flag = true;                tid = lenid;                while(k > 0){                    while(tid >= 0 && k < len[tid]){                        tid--;                    }                    if(k == len[tid] || k == len[tid+1] || k == len[tid]+1){                        if(k != len[tid]+1)flag = !flag;                        if(flag)ans = 0;                        else ans = 1;                        break;                    }                    k = len[tid+1] + 1 - k;                    flag = !flag;                }            }            printf("Case #%d: %lld\n",tcase,ans);        }    }    fclose(stdin);    fclose(stdout);    return 0;}

2.gCube 乘法越界，用log转换成加法，本题精度允许。

int main(){    int Tcase;    if(freopen("B-large-practice.in","r",stdin)==NULL)printf("failed!\n");    if(freopen("B-large-practice.out","w",stdout)==NULL)printf("failed!\n");    int len[1000];    int N,M,L,R;    double mul,high,low,mid,dim,tmppow;    while(scanf("%d",&Tcase) == 1){        for(int tcase = 1;tcase<= Tcase;tcase++){            scanf("%d %d",&N,&M);            for(int i = 0;i < N;i++){                scanf("%d",&len[i]);            }            printf("Case #%d:\n",tcase);            while(M--){                scanf("%d %d",&L,&R);                mul = 0;                for(int i =L;i<=R;i++)                        mul += log(len[i]);                high  = 1e9,low = 1;                dim = (R-L +1);                while(high-low > 1e-10){                    mid = (high + low )/ 2;                    tmppow =  dim*log(mid);                    if(mul-tmppow > 1e-10){                        low = mid;                    }else if(tmppow - mul > 1e-10){                        high = mid;                    }else break;                }                mid = (high+low) / 2;                printf("%.9lf\n",mid);            }        }    }    fclose(stdin);    fclose(stdout);    return 0;}

3.gCampus floyd全源最短路，保存原来存在的路径，如果新的最短路比原来存在的路短，那么就是原来的路可以被替换，本题有点恶心的是有若干起点终点重合的路径。

int main(){    int Tcase;    if(freopen("C-large-practice.in","r",stdin)==NULL)printf("failed!\n");    if(freopen("C-large-practice.out","w",stdout)==NULL)printf("failed!\n");    int N,M;    int oldpath[100][100];    int newpath[100][100];    vector<int> Pathid[100][100];    int U,V,C;    vector<int> res;    while(scanf("%d",&Tcase) == 1){        for(int tcase = 1;tcase<= Tcase;tcase++){            scanf("%d %d",&N,&M);            res.clear();            memset(oldpath,0,sizeof(oldpath));            memset(newpath,0,sizeof(newpath));            for(int i = 0;i < N;i++)for(int j = 0;j < N;j++)Pathid[i][j].clear();            for(int pathid = 0;pathid < M;pathid++){                scanf("%d %d %d",&U,&V,&C);                if(U==V){                        res.push_back(pathid);                        continue;                }else if(oldpath[U][V]!=0){                    if(C < oldpath[U][V]){                        for(int i = 0;i < Pathid[U][V].size();i++)                            res.push_back(Pathid[U][V][i]);                        Pathid[U][V].clear();                        Pathid[V][U].clear();                    }else if (C > oldpath[U][V]){                        res.push_back(pathid);                        continue;                    }else{                        Pathid[U][V].push_back(pathid);                        Pathid[V][U].push_back(pathid);                        continue;                    }                }                Pathid[U][V].push_back(pathid);                Pathid[V][U].push_back(pathid);                oldpath[U][V] =oldpath[V][U] = newpath[U][V]= newpath[V][U]=C;            }            for(int k = 0;k < N;k++){                for(int i = 0;i < N;i++){                    for(int j = 0;j < N;j++){                        if(i == j)continue;                        if(newpath[i][k] != 0 && newpath[k][j]!=0&&                           (newpath[i][j] == 0 ||newpath[i][k] + newpath[k][j] < newpath[i][j]))                            newpath[i][j] = newpath[j][i] = newpath[i][k] + newpath[k][j] ;                    }                }            }            printf("Case #%d:\n",tcase);            for(int i = 0;i < N;i++){                for(int j = i+1;j < N;j++){                        if(oldpath[i][j] != 0 && newpath[i][j] != oldpath[i][j]){                            for(int ii = 0;ii < Pathid[i][j].size();ii++)                                res.push_back(Pathid[i][j][ii]);                        }                }            }            sort(res.begin(),res.end());            for(int i = 0;i < res.size();i++)                printf("%d\n",res[i]);        }    }    fclose(stdin);    fclose(stdout);    return 0;}

4.gSnake 数据描述里的范围比较吓人，但是，实际时间长度最多也就X的上限+2倍的R或者L。

程序效率感觉有点低。

int main(){    int Tcase;    if(freopen("D-large-practice.in","r",stdin)==NULL)printf("failed!\n");    if(freopen("D-large-practice.out","w",stdout)==NULL)printf("failed!\n");    int S,R,C;    map<Point,int> flag;    set<Point> foodeaten;    queue<Point> snake;    CMD cmds[100000];    int cur_dir;    Point cur_point;    Point tmp_point;    int timenow;    int cmdid;    bool printflag;    while(scanf("%d",&Tcase) == 1){        for(int tcase = 1;tcase<= Tcase;tcase++){            scanf("%d %d %d",&S,&R,&C);            for(int i = 0;i < S;i++){                scanf("%d %c",&cmds[i].first,&cmds[i].second);            }            flag.clear();            foodeaten.clear();            while(!snake.empty())                snake.pop();            cur_dir = 0;//0 Right, 1 Down, 2 Left, 3 Up            cur_point = {0,0};//题目中是1,1开始，我这里自己调整到0,0            snake.push(cur_point);            flag[cur_point]++;            timenow = 1;            cmdid = 0;            printflag=true;            printf("Case #%d: ",tcase);            int counter = 0;            while(timenow <= 1e9 && counter <1000000){                if(cur_dir == 0){                    cur_point.second = (cur_point.second+1)% R ;                }else if(cur_dir == 1){                    cur_point.first = (cur_point.first+1) % C ;                }else if(cur_dir == 2){                    cur_point.second = (cur_point.second+R-1) % R;                }else{                    cur_point.first = (cur_point.first+C-1) % C ;                }                snake.push(cur_point);                if(((cur_point.first)%2) + ((cur_point.second)%2) == 1 && foodeaten.find(cur_point)==foodeaten.end()){                    foodeaten.insert(cur_point);                }else{                    tmp_point = snake.front();                    flag.erase(tmp_point);                    snake.pop();                }                flag[cur_point]++;                if(flag[cur_point] > 1){                    printf("%d\n",snake.size());                    printflag = false;                    break;                }                if(cmdid < S &&timenow == cmds[cmdid].first){                    if(cmds[cmdid].second == 'R')                        cur_dir = (cur_dir+1)%4;                    else                        cur_dir = (cur_dir+3)%4;                    cmdid++;                    counter=0;                }                timenow++,counter++;            }            if(printflag)printf("%d\n",snake.size());        }    }    fclose(stdin);    fclose(stdout);    return 0;}

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