A組C/C++下載鏈接：https://www.lanzous.com/i3jmx4b

B組C/C++下載鏈接：https://www.lanzous.com/i3jf1qj

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B組JAVA下載鏈接：https://www.lanzous.com/i3jmxli

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據(jù)官方消息：軟件類這周三會(huì)盡量出成績(jī)，最遲星期四。

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第一二題，基本nc題，代碼其實(shí)都不需要了。（我的答案僅供參考，畢竟一周后才出結(jié)果。祝愿各位都能有個(gè)好的成績(jī)！！！）

A題 組隊(duì)

490? ?

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B題 年號(hào)字串

BYQ

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C題 數(shù)列求值

【問(wèn)題描述】
給定數(shù)列1，1，1，3，5，9，17，..，從第4項(xiàng)開始，每項(xiàng)都是前3項(xiàng)的和。求第20190324項(xiàng)的最后4位數(shù)字。
【答案提交】
這是一道結(jié)果填空的題，你只需要算出結(jié)果后提交即可。

本題的結(jié)果為一個(gè)4位整數(shù)（提示：答案的千位不為0），在提交答案時(shí)只填寫這個(gè)整數(shù)，填寫多余的內(nèi)容將無(wú)法得分。

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#include <bits/stdc++.h>#define N 20190324 using namespace std;static int x = []() {std::ios::sync_with_stdio(false); cin.tie(0); return 0; }();int main() {vector<int> vec(N, 1);for (int i = 3; i < vec.size(); ++i) {vec[i] = vec[i - 1] + vec[i - 2] + vec[i - 3];if (vec[i] > 9999) {vec[i] = vec[i] - (vec[i] / 10000) * 10000;//只保留后四位，那我們就以10000為1個(gè)單位進(jìn)行減法，直到小于10000。 }}cout << vec[vec.size() - 1];return 0; }

?答案：4659

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D題 數(shù)的分解

【問(wèn)題描述】

把2019分解成3個(gè)各不相同的正整數(shù)之和，并且要求每個(gè)正整數(shù)都不包含數(shù)字2和4，一共有多少種不同的分解方法？.
注意交換3個(gè)整數(shù)的順序被視為同一種方法，例如1000-1001-18和1001+1000+18 被視 同一種。

哎，這是一個(gè)正整數(shù)，得從1開始，并且要想不重復(fù)，那么i,j,k三個(gè)數(shù)之間要有 i , j = i + 1, k = j + 1

#include <bits/stdc++.h>using namespace std;bool func(int n) {bool flag = true;while (1) {if (n < 10) {if (n == 2 || n == 4) {flag = false;break;}break;}int m = n % 10;n = n / 10;if (m == 2 || m == 4) {flag = false;break;}}return flag; }int main() {int ncount = 0;for (int i = 1; i < 2019; ++i) {for (int j = i + 1; j < 2019; ++j) {for (int k = j + 1; k < 2019; ++k) {if (func(i) && func(j) && func(k) && i + k + j == 2019) {ncount++;}}}}cout << ncount << endl;return 0; }

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?答案：40785

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?E題?迷宮

【問(wèn)題描述】
下圖給出了一個(gè)迷宮的平面圖，其中標(biāo)記為 1 的為障礙，標(biāo)記為 0 的為可以通行的地方。

010000
000100
001001
110000

迷宮的入口為左上角，出口為右下角，在迷宮中，只能從一個(gè)位置走到這個(gè)它的上、下、左、右四個(gè)方向之一。
對(duì)于上面的迷宮，從入口開始，可以按DRRURRDDDR 的順序通過(guò)迷宮，一共 10 步。其中 D、U、L、R 分別表示向下、向上、向左、向右走。
對(duì)于下面這個(gè)更復(fù)雜的迷宮（30 行 50 列） ，請(qǐng)找出一種通過(guò)迷宮的方式，其使用的步數(shù)最少，在步數(shù)最少的前提下，請(qǐng)找出字典序最小的一個(gè)作為答案。請(qǐng)注意在字典序中D<L<R<U。（如果你把以下文字復(fù)制到文本文件中，
請(qǐng)務(wù)必檢查復(fù)制的內(nèi)容是否與文檔中的一致。在試題目錄下有一個(gè)文件 maze.txt，內(nèi)容與下面的文本相同）

01010101001011001001010110010110100100001000101010
00001000100000101010010000100000001001100110100101
01111011010010001000001101001011100011000000010000
01000000001010100011010000101000001010101011001011
00011111000000101000010010100010100000101100000000
11001000110101000010101100011010011010101011110111
00011011010101001001001010000001000101001110000000
10100000101000100110101010111110011000010000111010
00111000001010100001100010000001000101001100001001
11000110100001110010001001010101010101010001101000
00010000100100000101001010101110100010101010000101
11100100101001001000010000010101010100100100010100
00000010000000101011001111010001100000101010100011
10101010011100001000011000010110011110110100001000
10101010100001101010100101000010100000111011101001
10000000101100010000101100101101001011100000000100
10101001000000010100100001000100000100011110101001
00101001010101101001010100011010101101110000110101
11001010000100001100000010100101000001000111000010
00001000110000110101101000000100101001001000011101
10100101000101000000001110110010110101101010100001
00101000010000110101010000100010001001000100010101
10100001000110010001000010101001010101011111010010
00000100101000000110010100101001000001000000000010
11010000001001110111001001000011101001011011101000
00000110100010001000100000001000011101000000110011
10101000101000100010001111100010101001010000001000
10000010100101001010110000000100101010001011101000
00111100001000010000000110111000000001000000001011
10000001100111010111010001000110111010101101111000

【答案提交】

這是一道結(jié)果填空的題，你只需要算出結(jié)果后提交即可。本題的結(jié)果為一個(gè)字符串，包含四種字母 D、U、L、R，在提交答案時(shí)只填寫這個(gè)字符串，填寫多余的內(nèi)容將無(wú)法得分。

#include <bits/stdc++.h>using namespace std;#define N 50 #define M 30int visited[M][N] = { 0 };//記憶已經(jīng)經(jīng)過(guò)的點(diǎn)（0表示未經(jīng)過(guò)，1表示經(jīng)過(guò)） int Move[4][2] = { {1,0},{0,-1},{0,1},{-1,0} };//用來(lái)移動(dòng)點(diǎn) char direction[4] = { 'D','L','R','U' };//與上面的二維數(shù)組的第一維索引號(hào)對(duì)應(yīng) struct Point {int x;//橫坐標(biāo)int y;//縱坐標(biāo)string str;//記憶點(diǎn)的軌跡int step;//經(jīng)過(guò)距離Point(int xx, int yy, int ss, string s) {//構(gòu)造函數(shù)x = xx;y = yy;str = s;step = ss;} };int MAP[M][N]{//迷宮 {0,1,0,1,0,1,0,1,0,0,1,0,1,1,0,0,1,0,0,1,0,1,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,0,0,0,1,0,0,0,1,0,1,0,1,0}, {0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,1,1,0,1,0,0,1,0,1}, {0,1,1,1,1,0,1,1,0,1,0,0,1,0,0,0,1,0,0,0,0,0,1,1,0,1,0,0,1,0,1,1,1,0,0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,0}, {0,1,0,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,1,0,0,1,0,1,1}, {0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,1,0,0,0,0,0,0,0,0}, {1,1,0,0,1,0,0,0,1,1,0,1,0,1,0,0,0,0,1,0,1,0,1,1,0,0,0,1,1,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,1,1,0,1,1,1}, {0,0,0,1,1,0,1,1,0,1,0,1,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,1,1,1,0,0,0,0,0,0,0}, {1,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,0,1,1,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0}, {0,0,1,1,1,0,0,0,0,0,1,0,1,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,1,1,0,0,0,0,1,0,0,1}, {1,1,0,0,0,1,1,0,1,0,0,0,0,1,1,1,0,0,1,0,0,0,1,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,1,0,1,0,0,0}, {0,0,0,1,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,1,0,0,1,0,1,0,1,0,1,1,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,1,0,1}, {1,1,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,1,0,0,1,0,0,0,1,0,1,0,0}, {0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,1,0,0,1,1,1,1,0,1,0,0,0,1,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,1}, {1,0,1,0,1,0,1,0,0,1,1,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,1,0,1,1,0,0,1,1,1,1,0,1,1,0,1,0,0,0,0,1,0,0,0}, {1,0,1,0,1,0,1,0,1,0,0,0,0,1,1,0,1,0,1,0,1,0,0,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,1,1,0,1,1,1,0,1,0,0,1}, {1,0,0,0,0,0,0,0,1,0,1,1,0,0,0,1,0,0,0,0,1,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,1,0,0}, {1,0,1,0,1,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,0,1,0,1,0,0,1}, {0,0,1,0,1,0,0,1,0,1,0,1,0,1,1,0,1,0,0,1,0,1,0,1,0,0,0,1,1,0,1,0,1,0,1,1,0,1,1,1,0,0,0,0,1,1,0,1,0,1}, {1,1,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,0,1,0}, {0,0,0,0,1,0,0,0,1,1,0,0,0,0,1,1,0,1,0,1,1,0,1,0,0,0,0,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,0,0,1,1,1,0,1}, {1,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,1,1,0,1,1,0,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,0,0,0,0,1}, {0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,1,0,1,0,1,0,1,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1}, {1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,1,0,0,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,0,1,0}, {0,0,0,0,0,1,0,0,1,0,1,0,0,0,0,0,0,1,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0}, {1,1,0,1,0,0,0,0,0,0,1,0,0,1,1,1,0,1,1,1,0,0,1,0,0,1,0,0,0,0,1,1,1,0,1,0,0,1,0,1,1,0,1,1,1,0,1,0,0,0}, {0,0,0,0,0,1,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,0,0,0,1,1,0,0,1,1}, {1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,1,1,1,1,0,0,0,1,0,1,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0}, {1,0,0,0,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,1,0,0,0,0,0,0,0,1,0,0,1,0,1,0,1,0,0,0,1,0,1,1,1,0,1,0,0,0}, {0,0,1,1,1,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,1,1}, {1,0,0,0,0,0,0,1,1,0,0,1,1,1,0,1,0,1,1,1,0,1,0,0,0,1,0,0,0,1,1,0,1,1,1,0,1,0,1,0,1,1,0,1,1,1,1,0,0,0} };bool check(int x, int y) {//檢查是否越界/走重復(fù)點(diǎn)/遇到墻 if (x < 0 || x >= M || y < 0 || y >= N || visited[x][y] || MAP[x][y] == 1)return false; return true; }void BFS() {queue<Point> q;Point p(0, 0, 0, "");//初始化隊(duì)列，從0,0點(diǎn)出發(fā) q.push(p);visited[0][0] = 1;//下面代碼的整體思路：將我們現(xiàn)在到達(dá)的點(diǎn)P周圍，可行點(diǎn)存入隊(duì)列中，選擇其中一個(gè)點(diǎn)作為點(diǎn)P更新，再將舊的P點(diǎn)從隊(duì)列中刪除，再以這個(gè)點(diǎn)重復(fù)之前找點(diǎn)的操作。//當(dāng)這條路為死路時(shí)，會(huì)回到隊(duì)列中另一個(gè)可行點(diǎn)，再次搜索，直到找到目標(biāo)點(diǎn)。while (!q.empty()) {Point fro = q.front();//取出隊(duì)頭if (fro.x == M - 1 && fro.y == N - 1) {//當(dāng)?shù)竭_(dá)右下角時(shí)，輸出路徑信息與步數(shù)，退出cout << fro.str << endl;cout << fro.step << endl;break;}q.pop();for (int i = 0; i < 4; ++i) {int nx = fro.x + Move[i][0];int ny = fro.y + Move[i][1];if (check(nx, ny)) {q.push(Point(nx, ny, fro.step + 1, fro.str + direction[i]));visited[nx][ny] = 1;}}} }int main() {BFS();return 0; }

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答案：DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR

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F題 特別數(shù)的和

小明對(duì)數(shù)位中含有 2、0、1、9 的數(shù)字很感興趣（不包括前導(dǎo)0），在1 到 40 中這樣的數(shù)包括1、2、9、10 至 32、39 和 40，共 28 個(gè)，他們的和是 574。請(qǐng)問(wèn)，在 1 到n 中，所有這樣的數(shù)的和是多少？

【輸入格式】
輸入一行包含兩個(gè)整數(shù)n。

【輸出格式】
輸出一行，包含一個(gè)整數(shù)，表示滿足條件的數(shù)的和。

【樣例輸入】

40

【樣例輸出】

574

【評(píng)測(cè)用例規(guī)模與約定】
對(duì)于20% 的評(píng)測(cè)用例，1≤n≤10
對(duì)于50% 的評(píng)測(cè)用例，1≤n≤100?
對(duì)于80% 的評(píng)測(cè)用例，1≤n≤1000
對(duì)于所有評(píng)測(cè)用例，1≤n≤10000?

#include <bits/stdc++.h>using namespace std;bool check(int n) {int flag = false;while (1) {if (n < 10) {if (n == 2 || n == 1 || n == 9 || n == 0) {flag = true;}break;}int k = n % 10;if (k == 2 || k == 1 || k == 9 || k == 0) {flag = true;}n /= 10;}return flag; }int main() {int num, sum = 0;cin >> num;for (int i = 1; i <= num; ++i) {if (check(i)) {sum += i;}}cout << sum;return 0; }

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G題 完全二叉樹的權(quán)值

給定一棵包含N 個(gè)節(jié)點(diǎn)的完全二叉樹，樹上每個(gè)節(jié)點(diǎn)都有一個(gè)權(quán)值，按從上到下、從左到右的順序依次是A1,A2,...,AN ，如下圖所示：

現(xiàn)在小明要把相同深度的節(jié)點(diǎn)的權(quán)值加在一起，他想知道哪個(gè)深度的節(jié)點(diǎn)權(quán)值之和最大？如果有多個(gè)深度的權(quán)值和同為最大，請(qǐng)你輸出其中最小的深度。
注：根的深度是1。

【輸入格式】
第一行包含一個(gè)整數(shù)N。
第二行包含N 個(gè)整數(shù)A1,A2,...,AN

【輸出格式】
輸出一個(gè)整數(shù)代表答案。

【樣例輸入】

7
1 6 5 4 3 2 1

【樣例輸出】

2

【評(píng)測(cè)用例規(guī)模與約定】
對(duì)于所有評(píng)測(cè)用例，1≤N≤100000 ，?100000≤Ai≤100000 。

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這道題，需要注意的地方是可以為負(fù)值，難度都不大，每個(gè)二叉樹每層最多2^(n-1)個(gè),所以，實(shí)際上就是比較第sum(1, 2^0),sum(2^0+1,2^1),sum(2^1+1,2^2)....中取max的深度n。

?我寫的這個(gè)復(fù)雜了，可以自己試試簡(jiǎn)化。

#include <bits/stdc++.h>using namespace std;int main() {int n, sum = 0, maxnum = 0, d = 1;cin >> n;vector<int> vec(n);for (int i = 0; i < n; ++i) {cin >> vec[i];}maxnum = vec[0];if (n == 1) {cout << d << endl;return 0;}vec.erase(vec.begin());for (int i = 1; pow(2, i) <= vec.size(); ++i) {sum = accumulate(vec.begin(), vec.begin() + pow(2, i), 0);//cout << sum << endl;//cout << sum << " " << maxnum << endl;if (sum > maxnum) {maxnum = sum;d = i + 1;}vec.erase(vec.begin(), vec.begin() + pow(2, i));}if (!vec.empty()) {sum = accumulate(vec.begin(), vec.end(), 0);if (sum > maxnum) {d = d + 1;}}cout << d;return 0; }

?

?H題 等差數(shù)列

數(shù)學(xué)老師給小明出了一道等差數(shù)列求和的題目。但是粗心的小明忘記了一部分的數(shù)列，只記得其中N 個(gè)整數(shù)。
現(xiàn)在給出這N 個(gè)整數(shù)，小明想知道包含這N 個(gè)整數(shù)的最短的等差數(shù)列有幾項(xiàng)？
【輸入格式】
輸入的第一行包含一個(gè)整數(shù)N。
第二行包含N 個(gè)整數(shù)A1,A2,...,AN (注意：A1~AN并不一定是按等差數(shù)列中的順序給出)

【輸出格式】
輸出一個(gè)整數(shù)表示答案。

【樣例輸入】

5
2 6 4 10 20

【樣例輸出】

10

【樣例說(shuō)明】
包含2、6、4、10、20 的最短的等差數(shù)列是2、4、6、8、10、12、14、16、18、20。

【評(píng)測(cè)用例規(guī)模與約定】
對(duì)于所有評(píng)測(cè)用例，2≤N≤100000 ，0≤Ai≤10^9

解決辦法是：排列數(shù)列之后，找出兩兩之間的差值最小值作為公差。這道難度不大，但是有坑在，常數(shù)列的數(shù)目是n ，所以要單獨(dú)討論常數(shù)列。

#include <bits/stdc++.h>using namespace std;int main() {int n, count;cin >> n;vector<int> vec(n), vec1(n - 1);for (int i = 0; i < n; ++i) {cin >> vec[i];}sort(vec.begin(), vec.end());//數(shù)列排序if (accumulate(vec.begin(), vec.end(), 0) == n * vec[0]) {//當(dāng)為常數(shù)列時(shí)，等差項(xiàng)有n項(xiàng)(前n項(xiàng)和==n*vec[0])cout << n;return 0;}for (int i = 0; i < n - 1; ++i) {vec1[i] = vec[i + 1] - vec[i];}sort(vec1.begin(), vec1.end());//數(shù)列之間從差，排序count = (vec.back() - vec.front()) / (*vec.begin()) + 1;cout << count;return 0; }

?

?

有N個(gè)“+”，M個(gè)“-”， N+M+1個(gè)整數(shù)

涼涼，坑啊！還以為25分這么好得，暴力解決，這道等結(jié)果吧，也不確定自己對(duì)沒有。

24個(gè)正號(hào)，1個(gè)負(fù)號(hào)，-25~-1 和 1這就26個(gè)整數(shù)，可以這樣計(jì)算1-(-1+(-2)+(-3)...+(-25))，要考慮括號(hào)。

測(cè)試用例：

2?1
1?-20?30?-100 答案是151，過(guò)程： 1+30-(-20+(-100))

?最后一道，我選擇戰(zhàn)略性放棄。

轉(zhuǎn)載于:https://www.cnblogs.com/Mayfly-nymph/p/10588986.html

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