签到 2016.3.6

1、CodeForces 58A Chat room

题意：
给定序列中是否存在“hello”这个子序列

解题思路：
O(n)扫一遍，逐个判断是否存在

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 100 + 5;int main(){    char s[maxn];    while (cin>>s) {        int len = strlen(s);        int Count = 0;        for (int i=0; i<len; ++i) {            if (Count == 0 && s[i] == 'h') {                ++Count;            }            if (Count == 1 && s[i] == 'e') {                ++Count;            }            if (Count == 3 && s[i] == 'l') {                ++Count;            }            if (Count == 2 && s[i] == 'l') {                ++Count;            }            if (Count == 4 && s[i] == 'o') {                ++Count;            }            if (Count == 5) {                break;            }        }        if (Count == 5) {            cout<<"YES"<<endl;        } else {            cout<<"NO"<<endl;        }    }    return 0;}

2、HDU 5630 Rikka with Chess
题意：
一个 n * m 的布满棋子的棋盘，只有黑白两种颜色，且相邻的棋子的颜色不一样
现在每次可以选择一个子矩形将其中所有棋子的颜色取反
问将棋盘上所有棋子的颜色都变成一样的最少操作次数

解题思路：
假设所有奇数行偶数列和偶数行奇数列的棋子都为黑色，其余都为白色
那么只需要将偶数行的棋子的颜色进行取反，然后再将偶数列的棋子的颜色取反
就可以以最少的操作次数将棋盘上所有棋子都变为白色

#include <iostream>using namespace std;int main(){    int T;    while (cin>>T) {        int n, m;        while (T--) {            cin>>n>>m;            cout<<n/2+m/2<<endl;        }    }    return 0;}

3、Codeforces_631A Interview
题意：
对于数组 x，f(x, L, R) = xL | x(L + 1) | … | xR
给出两个长度为 n 的数组 a，b（0 ≤ ai ≤ 10^9，0 ≤ bi ≤ 10^9）
找出两个整数 L，R （1 ≤ L ≤ R ≤ n）使得 f(a, L, R) + f(b, L, R) 最大
最后输出最大值

解题思路：
a | b >= a（a ≥ 0，b ≥ 0）

#include <iostream>#include <cstdio>using namespace std;int main(){    int n;    while (cin>>n) {        int i;        __int64 num1 = 0, num2 = 0;        int a[1010];        int b[1010];        for (i=0; i<n; ++i) {            cin>>a[i];            num1 |= a[i];        }        for (i=0; i<n; ++i) {            cin>>b[i];            num2 |= b[i];        }        cout<<num1+num2<<endl;    }    return 0;}

4、Codeforces_629A Far Relative’s Birthday Cake
题意：
一个 n*n 的正方形蛋糕是由许多个边长为 1 的正方形小块蛋糕组成
有些小块蛋糕上有巧克力
每一行和每一列的开心值为有巧克力的蛋糕块的对数
每一对只能被计数一次
求总的开心值

解题思路：
对于每块有巧克力的蛋糕，只将它与同一行左边和同一列上边的有巧克力的蛋糕配对
复杂度O(n^3)，但是 n 只有 1e3 ，所以想到啥写啥，懒得去再写O(n^2)的做法了，毕竟题有点水（虽然我也很水）

#include <iostream>using namespace std;int main(){    int n;    char s[110][110];    while (cin>>n) {        int i, j, k;        long long num = 0;        for (i=0; i<n; ++i) {            for (j=0; j<n; ++j) {                cin>>s[i][j];                if (s[i][j] == 'C') {                    for (k=0; k<j; ++k) {                        if (s[i][k] == 'C') {                            ++num;                        }                    }                    for (k=0; k<i; ++k) {                        if (s[k][j] == 'C') {                            ++num;                        }                    }                }            }        }        cout<<num<<endl;    }    return 0;}

5、 Codeforces_628A Tennis Tournament
题意：
n 个人两两进行比赛，只要输一次就会被淘汰，最后未淘汰的人数为 1 时结束比赛
每场比赛每名选手需要 b 瓶水，裁判需要 1 瓶水，为每名选手提供 p 条毛巾

解题思路：
比赛进行了 n - 1 场

#include <iostream>using namespace std;int main(){    int n, b, q;    while (cin>>n>>b>>q) {        int x, y;        x = (n-1)*(2*b+1);        y = n*q;        cout<<x<<" "<<y<<endl;    }    return 0;}

6、CodeForces_3A Shortest path of the king
题意：
给定起点和终点，每一步有 8 种走法，问最少几步能从起点走到终点
然后打印路径

解题思路：
1、根据终点对应起点的方位走
2、bfs

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned int uint;const int mod = 1e9 + 7;const int INF = 0x7fffffff;int dir[][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};int x1, x2, y_1, y2;bool vis[10][10];int pre[100][100];struct Node {    int x, y;    int Count;};int bfs(void);int main(){    char s[5], e[5];    cin >> s >> e;    x1 = 8 - (s[1] - '0');    x2 = 8 - (e[1] - '0');    y_1 = s[0] - 'a';    y2 = e[0] - 'a';    printf("%d\n", max(abs(x1-x2), abs(y_1-y2)));    if (x1 >= x2 && y_1 >= y2) {     //终点在起点的左上        while (!(x1 == x2 && y_1 == y2)) {            if (x1 > x2 && y_1 > y2) {                printf("LU\n");                --x1;                --y_1;            } else if (x1 > x2) {                printf("U\n");                --x1;            } else {                printf("L\n");                --y_1;            }        }    } else if (x1 >= x2 && y_1 <= y2) {      //终点在起点的右上        while (!(x1 == x2 && y_1 == y2)) {            if (x1 > x2 && y_1 < y2) {                printf("RU\n");                --x1;                ++y_1;            } else if (x1 > x2) {                printf("U\n");                --x1;            } else {                printf("R\n");                ++y_1;            }        }    } else if (x1 <= x2 && y_1 >= y2) {      //终点在起点的左下        while (!(x1 == x2 && y_1 == y2)) {            if (x1 < x2 && y_1 > y2) {                printf("LD\n");                ++x1;                --y_1;            } else if (x1 < x2) {                printf("D\n");                ++x1;            } else {                printf("L\n");                --y_1;            }        }    } else if (x1 <= x2 && y_1 <= y2) {      //终点在起点的右下        while (!(x1 == x2 && y_1 == y2)) {            if (x1 < x2 && y_1 < y2) {                printf("RD\n");                ++x1;                ++y_1;            } else if (x1 < x2) {                printf("D\n");                ++x1;            } else {                printf("R\n");                ++y_1;            }        }    }    return 0;}

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned int uint;const int mod = 1e9 + 7;const int INF = 0x7fffffff;int dir[][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};int x1, x2, y_1, y2;bool vis[10][10];int pre[100][100];struct Node {    int x, y;    int Count;};int bfs(void);int main(){    char s[5], e[5];    cin >> s >> e;    x1 = 8 - (s[1] - '0');    x2 = 8 - (e[1] - '0');    y_1 = s[0] - 'a';    y2 = e[0] - 'a';    printf("%d\n", bfs());    int x = x2, y = y2;    stack<string> S;    while (pre[x][y] != x*8 + y) {        int nx = pre[x][y] / 8;        int ny = pre[x][y] % 8;        string t;        if (ny < y) {            t += 'R';        }        if (ny > y) {            t += 'L';        }        if (nx > x) {            t += 'U';        }        if (nx < x) {            t += 'D';        }        S.push(t);        x = nx;        y = ny;    }    while (!S.empty()) {        cout << S.top() << endl;        S.pop();    }    return 0;}int bfs(void){    memset(vis, false, sizeof(vis));    Node node;    queue<Node> Q;    node.x = x1;    node.y = y_1;    node.Count = 0;    Q.push(node);    vis[x1][y_1] = true;    pre[x1][y_1] = x1*8 + y_1;    while (!Q.empty()) {        int nx = Q.front().x;        int ny = Q.front().y;        int t = nx*8 + ny;        int nCount = Q.front().Count;        if (nx == x2 && ny == y2) {            return nCount;        }        Q.pop();        for (int i = 0; i < 8; ++i) {            node.x = nx + dir[i][0];            node.y = ny + dir[i][1];            node.Count = nCount + 1;            if (0 <= node.x && node.x <= 8 && 0 <= node.y && node.y <= 8 && !vis[node.x][node.y]) {                vis[node.x][node.y] = true;                Q.push(node);                pre[node.x][node.y] = t;            }        }    }    return 0;}

7、CodeForces_4A Watermelon

题意：
判断一个数是否可以分解成 2 个非 0 的偶数

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned int uint;const int mod = 1e9 + 7;const int INF = 0x7fffffff;int main(){    int w;    scanf("%d", &w);    w -= 2;    if (w != 0 && (w&1) == 0) {        printf("YES\n");    } else {        printf("NO\n");    }    return 0;}

8、CodeForces_676A Nicholas and Permutation
题意：
n 个不同的整数 a1, a2, …, an (1 ≤ ai ≤ n)
把其中两个整数进行一次交换，使得 1 和 n 的距离最大
求这个最大的距离

解题思路：
将不改变 1 和 n 的位置、改变 1 的位置和改变 n 的位置后的三个距离进行比较，取最大值

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned int uint;const int mod = 1e9 + 7;const int INF = 0x7fffffff;const int maxn = 100 + 10;int a[maxn];int main(){    int n;    scanf("%d", &n);    int Min_pos, Max_pos;    for (int i = 1; i <= n; ++i) {        scanf("%d", &a[i]);        if (a[i] == 1) {            Min_pos = i;        }        if (a[i] == n) {            Max_pos = i;        }    }    if (Min_pos > Max_pos) {        swap(Min_pos, Max_pos);    }    int Max = Max_pos - Min_pos;    Max = max(Max, (Max_pos - 1));    Max = max(Max, (n - Min_pos));    printf("%d\n", Max);    return 0;}

9、CodeForces_677A Vanya and Fence
题意：
n 个人沿着高度为 h 的围墙走
为了不被警卫发现，比围墙高的人需要弯着腰
假设正常走路的人的宽度为 1，弯着腰走路的人的宽度为 2
问使他们不被发现的道路的最小的宽度

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned int uint;const double eps = 1e-8;const int INF = 0x7fffffff;const int maxn = 1000 + 10;int a[maxn];int main(){    int n, h;    scanf("%d%d", &n, &h);    int sum = 0;    for (int i = 0; i < n; ++i) {        scanf("%d", &a[i]);        if (a[i] > h) {            sum += 2;        } else {            sum += 1;        }    }    printf("%d\n", sum);    return 0;}

10、CodeForces_677B Vanya and Food Processor
题意：
给出 n 个土豆的高度、加工机器的高度和机器每秒钟可以粉碎的高度
问按顺序将这些土豆粉碎完需要多少时间

解题思路：
模拟
注意：开 long long 是一个好习惯，没开 long long 被 Hack 好桑心…

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <stack>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned int uint;const double eps = 1e-8;const int INF = 0x7fffffff;const int maxn = 1e5 + 10;int a[maxn];int main(){    int n, h, k;    scanf("%d%d%d", &n, &h, &k);    for (int i = 0; i < n; ++i) {        scanf("%d", &a[i]);    }    ll Count = 0;    int leave = 0;    int sum;    for (int i = 0; i < n; ++i) {        if (a[i] + leave > h) {            ++Count;            leave = 0;            --i;        } else {            sum = a[i] + leave;            int j = i + 1;            while (1) {                if (j >= n) {                    break;                }                if (sum + a[j] > h) {                    break;                }                sum += a[j];                ++j;                ++i;            }            if (sum < k) {                ++Count;                leave = 0;            } else {                Count += (sum/k);                leave = sum % k;            }        }    }    if (leave != 0) {        ++Count;    }    printf("%I64d\n", Count);    return 0;}