签到 2016.5.10

标签：

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 m 位，即求n mod 3^m??的三进制表示

复杂度 O(m)

#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;
}

3、HDU 5675 ztr loves math

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;
}

签到 2016.5.10

标签：

原文地址：http://blog.csdn.net/only_air/article/details/51359974