當前位置：首頁 > 编程资源 > 编程问答 >内容正文

编程问答

算法笔记实战模拟

發布時間：2025/3/21 编程问答 26 豆豆

生活随笔收集整理的這篇文章主要介紹了算法笔记实战模拟小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

文章目錄

A + B 和 C
部分 A+B
B1026 程序運行時間
B1046 劃拳
B1008 數組元素循環右移
數字圖像分類
B1018 錘頭剪刀布
shuffling Machine
Shortest Distance
B1010 一元多項式求導
A1009 多項式乘法
B1041 考試座位號
B1004成績排名
B1028 人口普查
World Cup Betting
A1006 sign in and sign out
A1036 boys vs girls
B1027 打印沙漏
A1031 Hello world for U
B1002 寫出這個數
B1014福爾摩斯的約會
B1024科學計數法
B1048數字加密

A + B 和 C

#include <iostream> #include <cstdio> #include <cstring>using namespace std;const int N = 80; int t, a, b, c;int main() {cin >> t;for (int i = 1; i <= t; i ++ ) {scanf("%d%d%d", &a, &b, &c);if (a + b > c) printf("Case #%d: true\n", i);else printf("Case #%d: false\n", i);}return 0; }

錯誤之處在于數據范圍爆炸了int，要用long long

#include <iostream> #include <cstdio> #include <cstring>using namespace std;const int N = 80; long long t, a, b, c;int main() {cin >> t;for (int i = 1; i <= t; i ++ ) {scanf("%lld%lld%lld", &a, &b, &c);if (a + b > c) printf("Case #%d: true\n", i);else printf("Case #%d: false\n", i);}return 0; }

int 范圍是 -2⁶³到2⁶³-1，爆了 int 用 long long

部分 A+B

#include <iostream> #include <cstdio> #include <cstring>using namespace std;typedef long long LL;const int N = 80;LL a, b, pa, pb; int da, db;int main() {cin >> a >> da >> b >> db;while (a) {int t = a % 10;if (t == da) {pa = 10 * pa + da;}a /= 10;}while (b) {int t = b % 10;if (t == db) {pb = 10 * pb + db;}b /= 10;}LL res = pa + pb;cout << res << endl;return 0; }

B1026 程序運行時間

近似函數round在cmath中

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 80;LL a, b; LL hh, mm, ss;int main() {cin >> a >> b;if (a > b) {LL t = a;a = b;b = t;}LL d = b - a;d = round(d / 100.0);hh = d / 3600;d %= 3600;mm = d / 60;d %= 60;ss = d;printf("%02lld:%02lld:%02lld\n", hh, mm, ss);return 0; }

問題在于輸出怎么填充0 %02d 整數，占兩位，不足填前導0

B1046 劃拳

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 110;int n; int aS, aD, bS, bD;int main() {cin >> n;int resA = 0, resB = 0;while (n -- ) {scanf("%d %d %d %d", &aS, &aD, &bS, &bD);if ((aD == aS + bS) && (bD != aD)) {resB ++ ;} else if ((bD == aS + bS) && (bD != aD)) {resA ++ ;}}cout << resA << ' ' << resB << endl;return 0; }

B1008 數組元素循環右移

錯誤代碼

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 110;int n, m; int a[2 * N];int main() {cin >> n >> m;for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);for (int i = n - 1; i >= 0; i -- ) a[i + m] = a[i];for (int i = 0; i < m; i ++ ) a[i] = a[n + i];for (int i = 0; i < n; i ++ ) {printf("%d", a[i]);if (i != n - 1) printf(" ");}printf("\n");return 0; }

錯誤原因 M 是可能大于 N 的
因此 m = n % m;

數字圖像分類

通過部分測評

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1010;int n; int a[10], d[N];int main() {while (scanf("%d", &n) != EOF) {bool flag = false; // 監測 A2 類int con = 1; // A2 類的運算符int cnt = 0; // 計算 A4 類的個數for (int i = 0; i < n ; i ++ ) scanf("%d", &d[i]);for (int i = 0; i < n ; i ++ ) {if ((d[i] % 5 == 0) && (d[i] % 2 == 0)) a[1] += d[i]; else if (d[i] % 5 == 1) {a[2] = a[2] + con * d[i];flag = true;con = -1 * con;}else if (d[i] % 5 == 2) a[3] ++ ;else if (d[i] % 5 == 3) {a[4] += d[i];cnt ++ ;}else a[5] = max(a[5], d[i]);}if (a[1]) printf("%d ", a[1]);else printf("N ");if (flag) printf("%d ", a[2]);else printf("N ");if (a[3]) printf("%d ", a[3]);else printf("N ");if (a[4]) printf("%.1f ", float(a[4]) / cnt);else printf("N ");if (a[5]) printf("%d\n", a[5]);else printf("N\n");}return 0; }

其實可以同時開一個計數數組和一個累加數值數組

B1018 錘頭剪刀布

簡單做法，沒有考慮字典序輸出

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e5 + 10;int n; int a[3], b[3]; int aCnt[3], bCnt[3];int main() {cin >> n;while (n -- ) {char l, r;cin >> l >> r;if ((l == 'J') && (r == 'B')) a[0] ++, b[2] ++ , aCnt[0] ++ ;if ((l == 'C') && (r == 'J')) a[0] ++, b[2] ++ , aCnt[1] ++ ;if ((l == 'B') && (r == 'C')) a[0] ++, b[2] ++ , aCnt[2] ++ ;if ((r == 'J') && (l == 'B')) b[0] ++, a[2] ++ , bCnt[0] ++ ;if ((r == 'C') && (l == 'J')) b[0] ++, a[2] ++ , bCnt[1] ++ ;if ((r == 'B') && (l == 'C')) b[0] ++, a[2] ++ , bCnt[2] ++ ;if ((r == 'J') && (r == l)) a[1] ++, b[1] ++;if ((r == 'C') && (r == l)) a[1] ++, b[1] ++;if ((r == 'B') && (r == l)) a[1] ++, b[1] ++;}printf("%d %d %d\n", a[0], a[1], a[2]);printf("%d %d %d\n", b[0], b[1], b[2]);if (aCnt[0] > aCnt[1] && aCnt[0] > aCnt[2]) printf("J ");else if (aCnt[1] > aCnt[0] && aCnt[1] > aCnt[2]) printf("C ");else printf("B ");if (bCnt[0] > bCnt[1] && bCnt[0] > bCnt[2]) printf("J\n");else if (bCnt[1] > bCnt[0] && bCnt[1] > bCnt[2]) printf("C\n");else printf("B\n");return 0; }

使用scanf()讀入c的時候有\n的問題，要使用 getchar 吸收掉換行符，同時可以做字母和數字之間的映射，減少復雜度。

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e5 + 10;int change(char c) {if (c == 'B') return 0;if (c == 'C') return 1;if (c == 'J') return 2; }int main() {char mp[3] = {'B', 'C', 'J'}; // mp[0] : 'B', mp[1] : 'C', mp[2] : 'J'int n;scanf("%d", &n);int time_A[3] = {0}, time_B[3] = {0}; // 記錄 A B 三種局勢的次數int hand_A[3] = {0}, hand_B[3] = {0}; // 記錄三種手勢獲勝的次數char c1, c2;int k1, k2;for (int i = 0; i < n; i ++ ) {getchar();scanf("%c %c", &c1, &c2);k1 = change(c1);k2 = change(c2);if ((k1 + 1) % 3 == k2) { // 甲獲勝time_A[0] ++ ;time_B[2] ++ ;hand_A[k1] ++ ;} else if (k1 == k2) { // 平局time_A[1] ++ ;time_B[1] ++ ;} else { // 乙獲勝time_A[2] ++ ;time_B[0] ++ ;hand_B[k2] ++ ;}}printf("%d %d %d\n", time_A[0], time_A[1], time_A[2]);printf("%d %d %d\n", time_B[0], time_B[1], time_B[2]);int id1 = 0, id2 = 0;for (int i = 0; i < 3; i ++ ) {if (hand_A[i] > hand_A[id1]) id1 = i;if (hand_B[i] > hand_B[id2]) id2 = i;}printf("%c %c\n", mp[id1], mp[id2]);return 0; }

標準答案版

shuffling Machine

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 60;int n; int go[N], back[N], form[N];int main() {cin >> n;char* mp[] = {"a", "S1", "S2", "S3", "S4", "S5", "S6", "S7", "S8", "S9", "S10", "S11", "S12", "S13","H1", "H2", "H3", "H4", "H5", "H6", "H7", "H8", "H9", "H10", "H11", "H12", "H13", "C1", "C2", "C3", "C4", "C5", "C6", "C7", "C8", "C9", "C10", "C11", "C12", "C13","D1", "D2", "D3", "D4", "D5", "D6", "D7", "D8", "D9", "D10", "D11", "D12", "D13", "J1", "J2"};for (int i = 1; i <= 54; i ++ ) scanf("%d", &form[i]), go[i] = i, back[i] = i;while (n -- ) {for (int i = 1; i <= 54; i ++ ) {back[form[i]] = go[i];}memcpy(go, back, sizeof(go));}for (int i = 1; i <= 54; i ++ ) {printf("%s", mp[go[i]]);if (i != 54) {printf(" ");}}printf("\n");return 0; }

是memcpy 而不是 memcopy 少一個字母 o
存放字符串用指針形變量申明方式是char * name[] =

Shortest Distance

我寫的代碼的問題，這份代碼超時

修改后

B1010 一元多項式求導

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1000 + 10;int m, n; double a[N], b[N]; bool st[N] = {false};int main() {cin >> m;int k;double x;while (m -- ) {scanf("%d %lf", &k, &x);a[k] = x;}cin >> n;while (n -- ) {scanf("%d %lf", &k, &x);b[k] = x;}int cnt = 0;for (int i = 0; i <= 1000; i ++ ) {if (a[i] != 0 || b[i] != 0) {a[i] += b[i];if (a[i] != 0) {st[i] = true;cnt ++ ;}}}printf("%d ", cnt);for (int i = 1000; i >= 0; i -- ) {if (st[i]) {printf("%d %.1lf", i, a[i]);if (cnt != 1) {printf(" ");}cnt -- ;}}printf("\n");return 0; }

A1009 多項式乘法

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 2000 + 10;struct pointMent {double res; // 答案double cof; // 系數 }p[N];int main() { int n, k;double x;cin >> n;for (int i = 0; i < n; i ++ ) {scanf("%d %lf", &k, &x);p[k].cof = x;}cin >> n;for (int i = 0; i < n; i ++ ) {scanf("%d %lf", &k, &x);for (int j = 0; j <= 1000; j ++ ) {p[j + k].res += p[j].cof * x; }}int cnt = 0;for (int i = 0; i < N; i ++ ) {if (p[i].res != 0.0) cnt ++ ;}printf("%d", cnt);for (int i = N; i >= 0; i -- ) {if (p[i].res != 0.0) {printf(" %d %.1f", i, p[i].res);}}printf("\n");return 0; }

double類型和0比較要用0.0比較，同一個位置要存儲多個信息的時候可以使用結構體

B1041 考試座位號

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1000 + 10;struct student {long long id; // 準考證考int test; // 試機號int exam; // 考試號 }stu[N];int n, m;int main() { cin >> n;for (int i = 0; i < n; i ++ ) {scanf("%lld %d %d", &stu[i].id, &stu[i].test, &stu[i].exam);}cin >> m;for (int i = 0; i < m; i ++ ) {int test;scanf("%d", &test);for (int j = 0; i < N; j ++ ) {if (stu[j].test == test) {printf("%lld %d", stu[j].id, stu[j].exam);if (i != m - 1) {printf("\n");}break;}}}printf("\n");return 0; }

使用結構體是關鍵

B1004成績排名

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1000 + 10;struct student {char name[20]; // 姓名char id[20]; // 學號int score; // 考試成績 }stu[N];int n;int main() { cin >> n;for (int i = 0; i < n; i ++ ) {scanf("%s %s %d", stu[i].name, stu[i].id, &stu[i].score);}int minId = 0;int maxId = 0;for (int i = 0; i < n; i ++ ) {if (stu[i].score > stu[maxId].score) maxId = i;if (stu[i].score < stu[minId].score) minId = i;}printf("%s %s\n", stu[maxId].name, stu[maxId].id);printf("%s %s\n", stu[minId].name, stu[minId].id);return 0; }

B1028 人口普查

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e5 + 10;struct student {char name[20]; // 姓名LL birthday; // 生日 }stu[N];int n;int main() { cin >> n;int cnt = 0;int minId = -1;int maxId = -1;for (int i = 0; i < n; i ++ ) {int yy, mm, dd;scanf("%s %d/%d/%d", stu[i].name, &yy, &mm, &dd);stu[i].birthday = yy * 10000 + mm * 100 + dd;if (stu[i].birthday >= 18140906 && stu[i].birthday <= 20140906) {cnt ++ ;if (stu[i].birthday > stu[minId].birthday || minId == -1) {minId = i;}if (stu[i].birthday < stu[maxId].birthday || maxId == -1) {maxId = i;}}}if (cnt) {printf("%d %s %s\n", cnt, stu[maxId].name, stu[minId].name);} else {printf("0\n"); }return 0; }

加上對所有都不合法的特判

World Cup Betting

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e5 + 10;struct student {char name[20]; // 姓名LL birthday; // 生日 }stu[N];int n;int main() { char mp[4] = {'W', 'T', 'L'};double p[4][4] = {};for (int i = 0; i < 3; i ++ ) {scanf("%lf %lf %lf", &p[i][0], &p[i][1], &p[i][2]);}double res = 1;for (int i = 0; i < 3; i ++ ) {int I = 0;double val = p[i][I];for (int j = 0; j < 3; j ++ ) {if (p[i][j] > val) {val = p[i][j];I = j;}}printf("%c ", mp[I]);res *= val;}res = (res * 0.65 - 1) * 2;printf("%.2lf\n", res);return 0; }

接收double數據類型一定要用%lf

A1006 sign in and sign out

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e5 + 10;struct person {char name[20]; // 姓名int start; // 進入int end; // 離開 }p[N];int n;int main() { cin >> n;int minId = 0;int maxId = 0;for (int i = 0; i < n; i ++ ) {int h1, m1, s1, h2, m2, s2;scanf("%s %d:%d:%d %d:%d:%d", p[i].name, &h1, &m1, &s1, &h2, &m2, &s2);p[i].start = h1 * 3600 + m1 * 60 + s1;p[i].end = h2 *3600 + m2 * 60 + s2;if (p[minId].start > p[i].start) {minId = i;}if (p[maxId].end < p[i].end) {maxId = i;}}printf("%s %s\n", p[minId].name, p[maxId].name);return 0; }

A1036 boys vs girls

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e5 + 10;struct person {char name[20]; // 姓名char sexy; // 性別char lesson[20]; // 課程int score; // 得分 }p[N];int n;int main() { cin >> n;int minId = -1;int maxId = -1;for (int i = 0; i < n; i ++ ) {scanf("%s %c %s %d", &p[i].name, &p[i].sexy, &p[i].lesson, &p[i].score);if (p[i].sexy == 'M' && (minId == -1 || p[minId].score > p[i].score)) {minId = i;}if (p[i].sexy == 'F' && (maxId == -1 || p[maxId].score < p[i].score)) {maxId = i;}}if (maxId == -1) {printf("Absent\n");} else {printf("%s %s\n", p[maxId].name, p[maxId].lesson);}if (minId == -1) {printf("Absent\n");} else {printf("%s %s\n", p[minId].name, p[minId].lesson);}if (maxId == -1 || minId == -1) {printf("NA\n");} else {printf("%d\n", p[maxId].score - p[minId].score);}return 0; }

B1027 打印沙漏

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e5 + 10;int n; char c;int main() { cin >> n >> c;int row = 1;int res = 0;while ((2 * row - 1 + 1) / 2 * row * 2 - 1 <= n ) {row ++ ;}row -- ;res = n - ((2 * row - 1 + 1) / 2 * row * 2 - 1);for (int i = 1; i <= row; i ++ ) {for (int j = i - 1; j > 0; j -- ) printf(" ");for (int j = 2 * (row - i + 1) - 1; j >= 1; j -- ) printf("%c", c);printf("\n");}for (int i = 2; i <= row; i ++ ) {for (int j = row - i; j > 0; j -- ) printf(" ");for (int j = 2 * i - 1; j > 0; j -- ) printf("%c", c);printf("\n");}printf("%d\n", res);return 0; }

A1031 Hello world for U

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 1e3;int n; char c[N][N], a[N];int main() { scanf("%s", a);int len = strlen(a);int side = (len + 2) / 3;int bottom = (len + 2) - side * 2;for (int i = 0; i < side; i ++ ) {c[i][0] = a[i];c[i][bottom - 1] = a[side + bottom + side - 3 - i];}for (int i = 0; i < bottom; i ++ ) {c[side - 1][i] = a[side - 1 + i];}for (int i = 0; i < side; i ++ ) {for (int j = 0; j < bottom; j ++ ) {if (c[i][j] != NULL) {printf("%c", c[i][j]);} else {printf(" ");}}printf("\n");}return 0; }

B1002 寫出這個數

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 110;char a[N]; char b[N][N]; int n, cnt[20];int main() { cin >> a;int len = strlen(a);char mp[][20] = {"lin", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};for (int i = 0; i < len; i ++ ) {cnt[a[i] - '0'] ++ ;}for (int i = 0; i <= 9; i ++ ) {n += i * cnt[i];}int num = 0;while (n) {int r = n % 10;strcpy(b[num], mp[r]);num ++ ;n /= 10;}for (int i = num - 1; i >= 0; i -- ) {printf("%s", b[i]);if (i != 0) {printf(" ");}}printf("\n");return 0; }

字符串復制strcpy(des, src)

B1014福爾摩斯的約會

錯誤

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 70;char a[10][N];int main() { char dayMap[][N] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};for (int i = 0; i < 4; i ++ ) {scanf("%s", a[i]);}int flag = 0;int len = strlen(a[0]);for (int i = 0; i < len; i ++ ) {if (a[0][i] == a[1][i] && (a[0][i] >= 'A' && a[0][i] <= 'G') && flag == 0) {printf("%s ", dayMap[a[0][i] - 'A']);flag ++ ;continue;}if (a[0][i] == a[1][i] && flag == 1) {if (a[0][i] >= '0' && a[0][i] <= 9) {printf("0%c:", a[0][i]);} else if (a[0][i] >= 'A' && a[0][i] <= 'N') {printf("%d:", a[0][i] - 'A' + 10);} else {continue;}flag ++ ;continue;}if (flag == 2) {break;}}len = strlen(a[2]);for (int i = 0; i < len; i ++ ) {if (a[2][i] == a[3][i] && ((a[2][i] >= 'a' && a[2][i] <= 'z') || (a[2][i] >= 'A' && a[2][i] <= 'Z'))) {printf("%02d\n", i);}}return 0; }

B1024科學計數法

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 10010;char str[N];int main() { scanf("%s", str);int len = strlen(str);if (str[0] == '-') printf("-");int pos = 0; // 存放 E 的位置while (str[pos] != 'E') {pos ++ ;}int exp = 0; // 存放指數for (int i = pos + 2; i < len; i ++ ) {exp = exp * 10 + (str[i] - '0');}if (exp == 0) {for (int i = 1; i < pos; i ++ ) {printf("%c", str[i]);}}if (str[pos + 1] == '-') { // 如果是負指數printf("0.");for (int i = 0; i < exp - 1; i ++ ) {printf("0");}printf("%c", str[1]); // 輸出除了小數點以外的數據for (int i = 3; i < pos; i ++ ) {printf("%c", str[i]);}} else { // 如果指數為正for (int i = 1; i < pos; i ++ ) { // 輸出小數點移動之后的數if (str[i] == '.') continue;printf("%c", str[i]);if (i == exp + 2 && pos - 3 != exp) {printf(".");}}// 如果指數 exp 較大，輸出多余的 0for (int i = 0; i < exp - (pos - 3); i ++ ) {printf("0");}}printf("\n");return 0; }

B1048數字加密

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h>using namespace std;typedef long long LL;const int N = 110;char A[N], B[N], ans[N]; char mp[20] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'J', 'Q', 'K'};void reverse(char s[]) {int len = strlen(s);for (int i = 0; i < len / 2; i ++ ) {char temp = s[i];s[i] = s[len - 1 - i];s[len - 1 - i] = temp;} }int main() { cin >> A >> B;reverse(A);reverse(B);int lenA = strlen(A);int lenB = strlen(B);int len = lenA >= lenB ? lenA : lenB;for (int i = 0; i < len; i ++ ) {char a = i < lenA ? A[i] : '0';char b = i < lenB ? B[i] : '0';if (i % 2 == 0) { // 奇數ans[i] = mp[(a - '0' + b - '0') % 13];} else { // 偶數ans[i] = (b - a + 10) % 10 + '0';}}reverse(ans);puts(ans);return 0; }

正確做法

總結

以上是生活随笔為你收集整理的算法笔记实战模拟的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。