数组（入门4）_vector <int> arr-优快云博客

本文链接：https://blog.youkuaiyun.com/2301_80161204/article/details/143987284

小鱼比可爱

#include <iostream>
#include <vector>
using namespace std;

int main() {

    int n;
    cin >> n;
    vector<int>arr(n);
    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }
    int count = 0;
    cout << "0" << " ";
    for (int i = 1; i < n; i++) {
        for (int j = i - 1; j >= 0; j--) {
            if (arr[i] > arr[j])count++;
        }
        cout << count << " ";
        count = 0;
    }

    return 0;
}

小鱼的数字游戏

#include <iostream>
#include <vector>
using namespace std;

int main() 
{
    vector<int>arr;
    int x;
    cin >> x;
    while (x) {
        arr.push_back(x);
        cin >> x;
    }
    for (int i = arr.size() - 1; i >= 0; i--) {
        cout << arr[i] << " ";
    }
    cout << endl;
    

    return 0;
}

【深基5.例3】冰雹猜想

#include <iostream>
#include <vector>
using namespace std;

int main() 
{
    int n;
    cin >> n;
    vector<int>arr;
    arr.push_back(n);
    while (n != 1) {
        if (n % 2 == 0) {
            n /= 2;
            arr.push_back(n);
        }
        else {
            n = 3 * n + 1;
            arr.push_back(n);
        }
    }
    for (int i = arr.size()-1; i >= 0; i--) {
        cout << arr[i] << " ";
    }

    return 0;
}

[NOIP2005 普及组] 校门外的树

#include <iostream>
#include <vector>
using namespace std;

int main() 
{
    int l, m;
    cin >> l >> m;
    
    vector<bool>trees(l + 1, true);
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        for (int j = u; j <= v; j++) {
            if (j >= 0 && j <= l) {
                trees[j] = false;
            }
        }
    }
    int remainTrees = 0;
    for (int i = 0; i < l + 1; i++) {
        if (trees[i])remainTrees++;
    }
    cout << remainTrees << endl;

    return 0;
}

注意：

像这种，有重合部分，而且如果分情况讨论的话，比较复杂多样，比如说这道题，需要判断第二次输入的u2是不是比上一个v1要小，是否相等，又或者比它大；如果是比较小，还要再判断v2是否也小于v1……

所以可以直接使用一个布尔向量，把所有在范围内的Tree都标记为false，同时只需要判断每一个点是否在有效范围内即可。

【深基5.例5】旗鼓相当的对手

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main() 
{
    int n;
    cin >> n;
    int sum1, sum2, count = 0;
    vector<vector<int>>score(n, vector<int>(3));
    for (int i = 0; i < n; i++) {
        cin >> score[i][0] >> score[i][1] >> score[i][2];
    }
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            sum1 = score[i][0] + score[i][1] + score[i][2];
            sum2 = score[j][0] + score[j][1] + score[j][2];
            int x = score[i][0] - score[j][0];
            int y = score[i][1] - score[j][1];
            int z = score[i][2] - score[j][2];
            if (abs(sum1 - sum2)<=10 &&
                abs(x)<=5 && 
                abs(y)<=5 && 
                abs(z)<=5) {
                count++;
            }
        }
    }
    cout << count << endl;

    return 0;
}

【深基5.例7】工艺品制作

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    int w, x, h;
    cin >> w >> x >> h;
    int q;
    cin >> q;
    int arr[20][20][20] = { 0 };
    while (q--) {
        int x1, y1, z1, x2, y2, z2;
        cin >> x1 >> y1 >> z1 >> x2 >> y2 >> z2;
        for (int i = x1 - 1; i < x2;i++ ){
            for (int j = y1 - 1; j < y2; j++) {
                for (int k = z1 - 1; k < z2; k++) {
                    arr[i][j][k] = 1;
                }
            }
        }
    }
    int count = 0;
    for (int i = 0; i < w; i++) {
        for (int j = 0; j < x; j++) {
            for (int k = 0; k < h; k++) {
                if (arr[i][j][k]==0) {
                    count++;
                }
            }
        }
    }
    cout << count << endl;

    return 0;
}

[AHOI2001] 彩票摇奖

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    int n;
    cin >> n;
    int a[7];
    vector<int>vec;
    for (int i = 0; i < 7; i++) {
        cin >> a[i];
    }
    int count = 0;
    for(int k=0;k<n;k++) {
        int x1, x2, x3, x4, x5, x6, x7;
        cin >> x1 >> x2 >> x3 >> x4 >> x5 >> x6 >> x7;
        for (int i = 0; i < 7; i++) {
            if (x1 == a[i])count++;
            if (x2 == a[i])count++;
            if (x3 == a[i])count++;
            if (x4 == a[i])count++;
            if (x5 == a[i])count++;
            if (x6 == a[i])count++;
            if (x7 == a[i])count++;
        }
        vec.push_back(count);
        count = 0;
    }
    for (int i = 7; i>0; i--) {
        for (int j = 0; j < n; j++) {
            if (vec[j] == i)count++;
        }
        cout << count << " ";
        count = 0;
    }

    return 0;
}

注意：

for(int k=0;k<n;k++)这一步不能用while(n--)来循环，因为这样的话，n会减少到0，等到后面需要判断vec[j]是否==i的时候，不会进入这个循环

整体思路：先用一个a[7]固定数组存放中奖号码，注意，只要有一样的数字就算有中奖号码，而不一定要一一对应。然后循环n次，找出每个彩票里面的中奖号码数，将其存放到vec数组中。最后只需要判断7~1个中奖号码的数量输出即可

[NOIP2015 提高组] 神奇的幻方

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    int n;
    cin >> n;
    vector<vector<int>>vec(n, vector<int>(n));
    vec[0][n/2] = 1;
    int r = 0, c =n/2;
    for (int i = 2; i <= n * n; i++) {
        if (r == 0 && c != n - 1) {
            vec[n - 1][c + 1] = i;
            r = n - 1, c += 1;
        }
        else if (c == n - 1 && r != 0) {
            vec[r - 1][0] = i;
            r -= 1, c = 0;
        }
        else if (r == 0 && c == n - 1) {
            vec[r + 1][c] = i;
            r += 1;
        }
        else if (r != 0 && c != n - 1) {
            if (vec[r - 1][c + 1] == 0
                && r-1>=0 && c+1<n) {
                vec[r - 1][c + 1] = i;
                r -= 1, c += 1;
            }
            else {
                vec[r + 1][c] = i;
                r += 1;
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << vec[i][j] << " ";
        }
        cout << endl;
    }

    return 0;
}

注意：

这道题我采用的是纯模拟，就是直接照着题目要求模拟就行，但是由于题目要求有点多，大家注意不要看差行了。

【深基5.例10】显示屏

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    int n;
    cin >> n;
    string ans[10];
    for (int i = 1; i <= n; i++) {
		char x;
		cin >> x;
		if (x == '1')
		{

			ans[1] += "..X.";
			ans[2] += "..X.";
			ans[3] += "..X.";
			ans[4] += "..X.";
			ans[5] += "..X.";
		}
		if (x == '2')
		{
			ans[1] += "XXX.";
			ans[2] += "..X.";
			ans[3] += "XXX.";
			ans[4] += "X...";
			ans[5] += "XXX.";
		}
		if (x == '3')
		{
			ans[1] += "XXX.";
			ans[2] += "..X.";
			ans[3] += "XXX.";
			ans[4] += "..X.";
			ans[5] += "XXX.";
		}
		if (x == '4')
		{
			ans[1] += "X.X.";
			ans[2] += "X.X.";
			ans[3] += "XXX.";
			ans[4] += "..X.";
			ans[5] += "..X.";
		}
		if (x == '5')
		{
			ans[1] += "XXX.";
			ans[2] += "X...";
			ans[3] += "XXX.";
			ans[4] += "..X.";
			ans[5] += "XXX.";
		}
		if (x == '6')
		{
			ans[1] += "XXX.";
			ans[2] += "X...";
			ans[3] += "XXX.";
			ans[4] += "X.X.";
			ans[5] += "XXX.";
		}
		if (x == '7')
		{
			ans[1] += "XXX.";
			ans[2] += "..X.";
			ans[3] += "..X.";
			ans[4] += "..X.";
			ans[5] += "..X.";
		}
		if (x == '8')
		{
			ans[1] += "XXX.";
			ans[2] += "X.X.";
			ans[3] += "XXX.";
			ans[4] += "X.X.";
			ans[5] += "XXX.";
		}
		if (x == '9')
		{
			ans[1] += "XXX.";
			ans[2] += "X.X.";
			ans[3] += "XXX.";
			ans[4] += "..X.";
			ans[5] += "XXX.";
		}
		if (x == '0')
		{
			ans[1] += "XXX.";
			ans[2] += "X.X.";
			ans[3] += "X.X.";
			ans[4] += "X.X.";
			ans[5] += "XXX.";
		}
    }
	for (int i = 1; i <= 5; i++) {
		ans[i] = ans[i].substr(0, ans[i].length()-1);
		cout << ans[i] << endl;
	}

    return 0;
}

注意：

这道题的话，感觉也没什么含金量（bushi），就是把依据输入的数字，一行一行的添加到String ans[10]（注意不是ans[5]这样的结果是错误的，因为我是从下标为1开始的，其实也可以设置为ans[6]）中。其次，就是ans[i] = ans[i].substr(0, ans[i].length()-1);注意是length-1，不是length

没必要用二维数组，感觉用二维数组反而麻烦

梦中的统计

#include <iostream>
#include <vector>
#include <string>
using namespace std;

int main()
{
    int m, n;
    cin >> m >> n;
    int arr[10] = { 0 };
    for (int i = m; i <= n; i++) {
        string s = to_string(i);
        for (int j = 0; j < s.length(); j++) {
            arr[s[j]-'0']++;
        }
    }
    for (int i = 0; i < 10; i++) {
        cout << arr[i] << " ";
    }

    return 0;
}

注意：

for循环嵌套时，一定要注意不能使用一样的变量名，我一开始习惯性全部设为i了，直接unaccept，泪目；还有就是没必要用很多if或者switch去判断每个s[j]，直接用s[j]-'0'就OK了，泪目，血的教训。上一道题“显示屏”我暂时没想到如果不用if或者switch的话还能用什么。

[NOIP2014 普及组] 珠心算测验

#include <iostream>
#include <unordered_set>
using namespace std;

const int M = 2e4 + 1;
int vis[M], a[102];
int n, ans;

int main()
{
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		vis[a[i]] = 1;
	}
	unordered_set<int>set;
	for (int i = 1; i <= n; i++) {
		for (int j = i + 1; j <= n; j++) {
			int sum = a[i] + a[j];
			if (vis[sum] == 1 &&
				a[i] != a[j] ) {
				set.insert(sum);
			}
		}
	}
	for (int x : set) {
		ans++;
	}
	cout << ans << endl;

	return 0;
}

注意：

ans不应该在set每次insert的时候++，因为有可能有重复的

爱与愁的心痛

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	int minn = 3e5 + 1;
	int n, m;
	cin >> n >> m;
	vector<int>vec(n);
	for (int i = 0; i < n; i++) {
		cin >> vec[i];
	}

	if (n == m) {
		int sum = 0;
		for (int i = 0; i < n; i++) {
			sum += vec[i];
		}
		cout << sum << endl;
		return 0;
	}

	for (int i = 0; i < n - m; i++) {
		int sum = 0;
		for (int j = i; j < i + m; j++) {
			sum += vec[j];
		}
		if (sum < minn)minn = sum;
	}
	cout << minn << endl;

	return 0;
}

注意：

这道题有两个坑点：

1，那个minn=3e5+1，不是3e3+1，我一开始直接按照m,n的最大值来设置的，结果ai<=100，还要再乘100，泪目

2，n==m的时候要单独判断

[USACO08OCT] Bovine Bones G

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	vector<int>pos(81);
	int a[4];
	for (int i = 1; i <= 3; i++)cin >> a[i];
	//按照筛子大小排序
	if (a[1] > a[2])swap(a[1], a[2]);
	if (a[2] > a[3])swap(a[2], a[3]);
	if (a[1] > a[2])swap(a[1], a[2]);
	//cout << a[1] << a[2] << a[3];
	for (int i = 1; i <= a[1]; i++) {
		for (int j = 1; j <= a[2]; j++) {
			for (int k = 1; k <= a[3]; k++) {
				int sum = i + j + k;
				pos[sum]++;
			}
		}
	}
	int max_count = 0;
	int ans = 0;
	for (int i = 3; i < a[1] + a[2] + a[3]; i++) {
		if (pos[i] > max_count) {
			max_count = pos[i];
			ans = i;
		}
		else if (pos[i] == max_count && i < ans) {
			ans = i;
		}
	}
	cout << ans << endl;

	return 0;
}

注意：

最好定义两个变量，一个max_count用来记录pos[i]，一个ans用来记录i，这样不会混，我一开始就是自己都搞混了

开灯

#include <iostream>
#include <vector>
using namespace std;

const int m = 2e6 + 1;

int main()
{
	int n;
	cin >> n;
	vector<bool>vec(m,false);
	for (int i = 0; i < n; i++) {
		double a;
		int t;
		cin >> a >> t;
		for (int j = 1; j <= t; j++) {
			if (vec[int(j * a)])vec[int(j * a)] = false;
			else vec[int(j * a)] = true;
		}
	}
	int count = 0;
	for (int i = 0; i < m; i++) {
		if (vec[i]) {
			cout << i << endl;
			break;
		}
	}

	return 0;
}

【深基5.习6】蛇形方阵

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	int n;
	cin >> n;
	vector<vector<int>>vec(n, vector<int>(n));
	int startX = 0, startY = 0;
	int endX = n - 1, endY = n - 1;
	int loop = n / 2;
	int count = 1;
	while (loop--) {
		for (int j = startY; j < endY; j++) {
			vec[startX][j] = count++;
		}
		for (int i = startX; i < endX; i++) {
			vec[i][endY] = count++;
		}
		for (int j = endY; j > startY; j--) {
			vec[endX][j] = count++;
		}
		for (int i = endX; i > startX; i--) {
			vec[i][startY] = count++;
		}
		startX++;
		startY++;
		endY--;
		endX--;
	}
	if (n % 2 == 1)vec[n / 2 ][n / 2 ] = count;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			printf("%3d", vec[i][j]);
		}
		cout << endl;
	}

	return 0;
}

注意：

如果是奇数的话，应该把vec[n/2][n/2]的位置填充为count，注意不是n/2+1，因为数组下标从0开始

【深基5.习7】杨辉三角

#include <iostream>
using namespace std;

int a[20][20];

int main()
{
	int n;
	cin >> n;
	a[0][0] = 1;
	for (int i = 1; i < 20; i++) {
		a[i][0] = 1;
		a[i][i] = 1;
	}

	for (int i = 2; i < 20; i++) {
		for (int j = 1; j <= i; j++) {
			a[i][j] = a[i - 1][j - 1] + a[i - 1][j];
		}
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j <= i; j++) {
			cout << a[i][j] << " ";
		}
		cout << endl;
	}


	return 0;
}

注意：

第二个for循环的嵌套循环中，j<=i，我一开始没注意写成了j<19,；还有输出的时候的i<n，j<=i

【Mc生存】插火把

#include <iostream>
using namespace std;

int a[102][102];

int main()
{
	
	int n, m, k;
	cin >> n >> m >> k;
	while (m--) {
		int x, y;
		cin >> x >> y;
		for (int i = x - 2; i <= x + 2; i++) {
			if (i > 0 && i<101)a[i][y] = 1;
		}
		for (int j = y - 2; j <= y + 2; j++) {
			if (j > 0 && j < 101)a[x][j] = 1;
		}
		if (x - 1 > 0 && y-1 > 0)a[x - 1][y - 1] = 1;
		if (x - 1 > 0 && y + 1 < 101)a[x - 1][y + 1] = 1;
		if (x + 1 < 101 && y - 1 > 0)a[x + 1][y - 1] = 1;
		if (x + 1 < 101 && y + 1 < 101)a[x + 1][y + 1] = 1;
	}
	while (k--) {
		int x, y;
		cin >> x >> y;
		for (int i = x - 2; i <= x + 2; i++) {
			for (int j = y - 2; j <= y + 2; j++) {
				if(i>0 && i<101 &&
					j>0 && j<101)
					a[i][j] = 1;
			}
		}
	}
	int count = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (a[i][j] == 0)count++;
		}
	}
	cout<<count<<endl;

	return 0;
}

压缩技术

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	int n;
	cin >> n;
	vector<vector<int>>vec(n, vector<int>(n));
	int count = 0;
	int right = 1;
	int sum = 0;
	while (sum!=n*n) {
		int x;
		cin >> x;
		if (right>0) {
			for (int i = 0; i < x; i++) {
				if (count == n) {
					cout << endl;
					count = 0;
				}
				cout << "0";
				count++;
			}
		}
		else {
			for (int i = 0; i < x; i++) {
				if (count == n) {
					cout << endl;
					count = 0;
				}
				cout << "1";
				count++;
			}
		}
		right = -right;
		sum += x;
	}



	return 0;
}

压缩技术（续集版）

#include <iostream>
using namespace std;

int main()
{
	string a[201];
	char c = '0';
	int sum = 0;
	cin >> a[0];
	for (int i = 1; i < a[0].size(); i++) {
		cin >> a[i];
	}
	cout << a[0].size() << " ";
	for (int i = 0; i < a[0].size(); i++) {
		for (int j = 0; j < a[0].size(); j++) {
			if (a[i][j] == c)sum++;
			else {
				c = a[i][j];
				cout << sum << " ";
				sum = 0;
				sum++;
			}
		}
	}
	cout << sum << endl;

	return 0;
}

注意：

第一个for循环的初始值是i=1

[USACO1.2] 方块转换 Transformations

#include <iostream>
using namespace std;

char a[15][15], b[15][15], c[15][15], d[15][15];
int n;

bool rotate90()
{
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			b[j][n - i + 1] = a[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (b[i][j] != c[i][j]) {
				return false;
			}
		}
	}
	return true;
}

bool rotate180()
{
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			b[n - i + 1][n - j + 1] = a[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (b[i][j] != c[i][j]) {
				return false;
			}
		}
	}
	return true;
}

bool rotate270()
{
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			b[n - j + 1][i] = a[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (b[i][j] != c[i][j]) {
				return false;
			}
		}
	}
	return true;
}

bool reflect()
{
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			b[i][n - j + 1] = a[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (b[i][j] != c[i][j]) {
				return false;
			}
		}
	}
	return true;
}

bool mix()
{
	reflect();
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			a[i][j] = b[i][j];
		}//更新数组a，因为经过了一次reflect函数
	}
	if (rotate90())return true;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			a[i][j] = b[i][j];
		}
	}
	if (rotate180())return true;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			a[i][j] = b[i][j];
		}
	}
	if (rotate270())return true;
	return false;
}

bool same()
{
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (b[i][j] != c[i][j]) {
				return false;
			}
		}
	}
	return true;
}

void solve()
{
	if (rotate90()) {
		cout << "1" << endl;
		return;
	}
	if (rotate180()) {
		cout << "2" << endl;
		return;
	}
	if (rotate270()) {
		cout << "3" << endl;
		return;
	}
	if (reflect()) {
		cout << "4" << endl;
		return;
	}
	if (mix()) {
		cout << "5" << endl;
		return;
	}
	if (same()) {
		cout << "6" << endl;
		return;
	}
	cout << "7" << endl;
	return;
}

int main()
{
	cin >> n;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> a[i][j];
			d[i][j] = a[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> c[i][j];
		}
	}
	solve();

	return 0;
}