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最新推荐文章于 2018-05-10 13:25:38 发布

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最新推荐文章于 2018-05-10 13:25:38 发布

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1、HDU 5670 Machine

解题思路：

变化的规律为红 -> 绿 -> 蓝 -> 红

将红、绿、蓝分别表示0、1、2

原问题就转化为求 n 的三进制表示的最低的位，即求 mod $3^m$ 的三进制表示

复杂度

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long LL;

const int maxn = 30 + 5;
char s[maxn];

int m;
LL n;

int main()
{
//    freopen("in.txt", "r", stdin);
    int T;
    while (scanf("%d", &T) != EOF) {
        while (T--) {
            cin>>m>>n;
            for (int i=0; i<m; ++i) {
                s[i] = 'R';
            }
            LL x = n%3;
            LL y = n/3;
            for (int i=m-1; i>=0; --i) {
                if (x == 1) {
                    s[i] = 'G';
                } else if (x == 2) {
                    s[i] = 'B';
                }
                if (y == 0) {
                    break;
                }
                x = y%3;
                y /= 3;
            }
            for (int i=0; i<m; ++i) {
                cout<<s[i];
            }
            cout<<endl;
        }
    }
    return 0;
}

2、HDU 5671 Matrix

解题思路：

暴力肯定是不可行的

可以用两个数组记录行和列交换的操作

再用两个数组记录行和列加一个数的操作

复杂度 O(mn + q)

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 1000 + 10;
int Matrix[maxn][maxn];
int row[maxn], row_add[maxn];       //行
int column[maxn], column_add[maxn]; //列

int main()
{
//    freopen("in.txt", "r", stdin);
    int T;

    while (scanf("%d", &T) != EOF) {
        while (T--) {
            int n, m, q;
            scanf("%d%d%d", &n, &m, &q);
            for (int i=1; i<=n; ++i) {
                for (int j=1; j<=m; ++j) {
                    scanf("%d", &Matrix[i][j]);
                }
            }
            int a, x, y;
            memset(row_add, 0, sizeof(row_add));
            memset(column_add, 0, sizeof(column_add));
            for (int i=1; i<=n; ++i) {
                row[i] = i;
            }
            for (int i=1; i<=m; ++i) {
                column[i] = i;
            }
            for (int i=0; i<q; ++i) {
                scanf("%d%d%d", &a, &x, &y);
                if (a == 1) {
                    int t = row[x];
                    row[x] = row[y];
                    row[y] = t;
                } else if (a == 2) {
                    int t = column[x];
                    column[x] = column[y];
                    column[y] = t;
                } else if (a == 3) {
                    row_add[row[x]] += y;
                } else {
                    column_add[column[x]] += y;
                }
            }
            int t;
            for (int i=1; i<=n; ++i) {
                for (int j=1; j<m; ++j) {
                    t = Matrix[row[i]][column[j]] + row_add[row[i]] + column_add[column[j]];
                    printf("%d ", t);
                }
                t = Matrix[row[i]][column[m]] + row_add[row[i]] + column_add[column[m]];
                printf("%d\n", t);
            }
        }
    }
    return 0;
}

题意：
是否存在正整数 x 和 y，满足 n = x^2 - y^2 （n 为正整数）

解题思路：
据说这次 BC 出现很多问题（导致unrated）所以肯定不知道哪里有毒
一：
n = x^2 - y^2 = (x+y) * (x-y)

令 a = x+y; b = x-y;
那么 n = a*b; x = (a+b) / 2; y = (a-b) / 2;

所以 (a-b) != 0 && (a-b) %2 == 0 时有解

复杂度O(sqrt(n))

以为会TLE，没想到358MS过了

二：
假设存在解时 y = x+i; n = (x+i)^2 - x^2;
那么 n = 2xi + i^2;

i = 1时，n = 2x + 1;
i = 2时，n = 4x + 4;

容易得出结论，当 n 不等于 1 && n 不等于 4 && n 为奇数或者 4 的倍数时，方程一定有正整数解

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

typedef long long LL;

int main()
{
//    freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        LL n;
        scanf("%I64d", &n);
		bool flag = false;
		int t = (int)sqrt(n);
		for (int i=1; i<=t; ++i) {
			if (n%i == 0) {
				LL a = i;
				LL b = n / i;
				if ((a-b) != 0 && (a-b)%2 == 0) {
//                            cout<<a<<" "<<b<<endl;
					flag = true;
					break;
				}
			}
		}
		if (flag) {
			printf("True\n");
		} else {
			printf("False\n");
		}
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

typedef long long LL;

int main()
{
//    freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        LL n;
        scanf("%I64d", &n);
        if (n != 1 && n != 4 && (n%2 != 0 || n%4 == 0)) {
            printf("True\n");
        } else {
            printf("False\n");
        }
    }
    return 0;
}

4、CodeForces_673A Bear and Game

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>

using namespace std;

typedef long long LL;

const int maxn = 100;
int t[maxn];

int main()
{
#ifdef __AiR_H
    freopen("in.txt", "r", stdin);
#endif // __AiR_H
    int n;
    scanf("%d", &n);
    t[0] = 0;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &t[i]);
    }
    ++n;
    t[n] = 90;
    bool flag = false;
    for (int i = 1; i <= n; ++i) {
        if (t[i] - t[i-1] > 15) {
            flag = true;
            printf("%d\n", t[i-1] + 15);
            break;
        }
    }
    if (!flag) {
        printf("90\n");
    }
    return 0;
}

5、CodeForces_673B Problems for Round

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

typedef long long LL;

const int maxn = 1e5 + 10;

vector<int> Div1, Div2;

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    int u, v;
    for (int i = 0; i < m; ++i) {
        scanf("%d%d", &u, &v);
        if (u < v) {
            swap(u, v);
        }
        Div1.push_back(u);
        Div2.push_back(v);
    }
    if (m == 0) {
        printf("%d\n", n-1);
    } else {
        int a = *max_element(Div2.begin(), Div2.end());
        int b = *min_element(Div1.begin(), Div1.end());
        printf("%d\n", max(0, b-a));
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

typedef long long LL;

const int maxn = 1e5 + 10;

int main()
{
#ifdef __AiR_H
    freopen("in.txt", "r", stdin);
#endif // __AiR_H
    int n, m;
    scanf("%d%d", &n, &m);
    int u, v;
    int div2_Max = 1, div1_Min = n;
    for (int i = 0; i < m; ++i) {
        scanf("%d%d", &u, &v);
        if (u < v) {
            swap(u, v);
        }
        div1_Min = min(div1_Min, u);
        div2_Max = max(div2_Max, v);
    }
    if (m == 0) {
        printf("%d\n", n-1);
    } else {
        printf("%d\n", max(0, div1_Min - div2_Max));
    }
    return 0;
}

6、HDU 2054 A == B ?

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 1e5 + 10;
char A[maxn], B[maxn];

void Init(char s[]);

int main()
{
#ifdef __AiR_H
    freopen("in.txt", "r", stdin);
#endif // __AiR_H
    while (scanf("%s%s", A, B) != EOF) {
        Init(A);
        Init(B);
        if (strcmp(A, B) == 0) {
            printf("YES\n");
        } else {
            printf("NO\n");
        }
    }
    return 0;
}

void Init(char s[])
{
    int len = strlen(s);
    //strstr(str1,str2)
    //若str2是str1的子串，则返回str2在str1的首次出现的地址
    //如果str2不是str1的子串，则返回NULL
    if (strstr(s, ".")) {
        while (s[len-1] == '0') {
            s[len-1] = '\0';
            --len;
        }
        if (s[len-1] == '.') {
            s[len-1] = '\0';
        }
    }
}

7、Codeforces_5A Chat Server's Outgoing Traffic

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 100 + 10;
char s[maxn];

int main()
{
#ifdef __AiR_H
    freopen("in.txt", "r", stdin);
#endif // __AiR_H
    int Count = 0;
    int ans = 0;
    while (gets(s)) {
        if (s[0] == '+') {
            ++Count;
        } else if (s[0] == '-') {
            --Count;
        } else {
            int len = strlen(s);
            int pos = 0;
            while (s[pos] != ':') {
                ++pos;
            }
            ans += Count * (len - (pos+1));
        }
    }
    printf("%d\n", ans);
    return 0;
}

8、Codeforces_686A Free Ice Cream

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
char cmd[2];

int main()
{
#ifdef __AiR_H
    freopen("in.txt", "r", stdin);
#endif // __AiR_H
    ull n, x;
    ull d;
    scanf("%I64d%I64d", &n, &x);
    ull Count = 0;
    for (ull i = 0; i < n; ++i) {
        scanf("%s%I64d", cmd, &d);
        if (cmd[0] == '+') {
            x += d;
        } else {
            if (x < d) {
                ++Count;
            } else {
                x -= d;
            }
        }
    }
    printf("%I64d %I64d\n", x, Count);
    return 0;
}

9、Codeforces_686B Little Robber Girl's Zoo

题目要求操作次数不能超过 20000 ，n^2 的循环肯定会 wa，但是不确定 n^3 循环操作次数会不会超 20000

不过最后 AC 了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 100 + 10;
int a[maxn];

int main()
{
#ifdef __AiR_H
    freopen("in.txt", "r", stdin);
#endif // __AiR_H
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    int s = 0, e = 0;
    for (int i = 1; i <= n*n; ++i) {
        bool flag = false;
        for (int j = n; j > 0; --j) {
            if (a[j] < a[j-1]) {
                if (!flag) {
                    flag = true;
                    e = j;
                }
                --j;
                s = j;
            } else {
                if (flag) {
                    printf("%d %d\n", s, e);
                    flag = false;
                    for (int k = s; k <= e; k+=2) {
                        swap(a[k], a[k+1]);
                    }
                }
            }
        }
        if (flag) {
            printf("%d %d\n", s, e);
            for (int j = s; j <= e; j+=2) {
                swap(a[j], a[j+1]);
            }
        }
    }
    return 0;
}

10、HDU 1008 Elevator

解题思路：

注意有可能在同一层停留

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 100 + 10;
int num[maxn];

int main()
{
#ifdef __AiR_H
    freopen("in.txt", "r", stdin);
#endif // __AiR_H
    int N;
    while (scanf("%d", &N) != EOF && N != 0) {
        int ans = 0;
        num[0] = 0;
        for (int i = 1; i <= N; ++i) {
            scanf("%d", &num[i]);
            if (num[i-1] < num[i]) {
                ans += (num[i] - num[i-1]) * 6 + 5;
            } else {
                ans += (num[i-1] - num[i]) * 4 + 5;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}