1-2 暴力破解-模拟

模拟：根据题目要求编写代码
可分为：图形排版（根据某种规则输出特定图形）、日期问题、其他模拟

一.图形排版
1.输出梯形（清华大学）

法一：等差数列
分析：每行的星号个数为等差数列2n+2（n=1,2,3,…）

#include<iostream>
using namespace std;
int main() {int n = 0;cin >> n;for (int i = 1; i <= n; i++) {int j = 0;for (; j < 2 * n + 2 - (2 * i + 2); j++) {cout << " ";}for (; j < 2 * n + 2; j++) {cout << "*";}cout << endl;}
}

法二：梯形求解
分析：设第0行有h个星号，第1行有h+2个，第2行有h+2* 2个，第i行有h+2* i个
若该梯形的高为h，则最后一行是第h-1行，星号数为h+2* (h-1)
则第一行的空格数=h+2* (h-1)-h
第i行的空格数=h+2* (h-1)-(h+2* i)=2(h-1-i)个

即第i行的空格数为2(h-1-i)个，星号数为h+2* i个

#include<iostream>
using namespace std;
int main() {int h=0;while (scanf("%d",&h) != EOF) {for (int i = 0; i < h; i++) {for (int j = 0; j < 2 * (h - 1 - i); j++) {cout << " ";}for (int j = 0; j < h+2*i; j++) {cout << "*";}cout << endl;}}
}

2.叠筐（杭州电子科技大学-2074）

评测系统

法一：

#include<iostream>
using namespace std;
int main() {int n=0;char b, a;while (cin >> n) {cin >> b >> a;string out[100];int count = 0;//特殊处理if (n == 1) {cout << b <<endl << endl;continue;}//决定开头字符if ((n - 1) / 2 % 2 == 0) {int temp = a;a = b;b = temp;}//第一行out[count] += " ";for (int i = 0; i < n - 2; i++) {out[count] += a;}out[count] += " ";count++;//第二行往后int num = 0;char arr[100] = { 0 };for (int i = 0; i < n / 2; i++) {if (i) {int temp = a;a = b;b = temp;}arr[num] = a;arr[n - num] = a;for (int j = 0; j <= num; j++) {out[count] += arr[j];}for (int j = num + 1; j < n - num - 1; j++) {out[count] += b;}for (int j = n - num; j <= n; j++) {out[count] += arr[j];}num++;count++;}//输出for (int i = 0; i < count; i++) {cout << out[i] << endl;}for (int i = count - 2; i >= 0; i--) {cout << out[i] << endl;}cout << endl;}
}

法二：
先补全四个角，可以看出，这是一个同心正方形，边长为n

以最外层左上角为坐标原点(0,0)，向下向右展开坐标轴，右下角坐标为(n-1,n-1)，最外层正方形边长为n
对于任意内层正方形，设左上角的坐标为(i,i)，右下角的坐标为(j,j)，则该正方形边长为n-2i
(i,i)和(j,j)两坐标相加即可得到右下角的坐标(n-1,n-1)，因此有等式i+j=n-1成立

i的取值从0到n/2，当i=n/2时为中心花色

设输入的第一个字符是B，第二个字符是A

中心花色是第一个输入字符

向外圈填充，距离中心距离为1填充字符为第二个输入字符

向外圈填充，距离为2填充第一个输入字符

以此类推，左上角坐标(i,i)距离中心的距离=n/2-i
距离为奇数填充第二个输入字符
距离为偶数填充第一个输入字符

char c;//当前圈层填充的字符
if ((n / 2 - i) % 2 == 0) {//偶数c = 第一个输入字符;
}
else {c = 第二个输入字符;
}

完整代码

#include<iostream>
using namespace std;
char matrix[80][80];//输出数组
int main() {int n;char a, b;bool firstCase = true;while (scanf("%d %c %c", &n, &a, &b) != EOF) {if (firstCase == true) {firstCase = false;//第一组前面无需换行}else { //在运行前先输出换行，确保两组用例之间有换行printf("\n");}for (int i = 0; i <= n / 2; i++) {//(i,i)为某正方形左上角坐标int j = n - 1 - i;//(j,j)为某正方形右下角坐标int length = n - 2 * i;//正方形边长char c;//当前圈层填充的字符if ((n / 2 - i) % 2 == 0) {//偶数c = a;}else {c = b;}for (int k = 0; k < length; k++) {//为当前圈层赋值matrix[i][i + k] = c;//上边赋值matrix[i+k][i] = c;//左边赋值matrix[j][j-k] = c;//下边赋值matrix[j-k][j] = c;//右边赋值}}if (n != 1) {//剔除四个角，确保外框尺寸n=1时也有输出，此时输出的是中心花色matrix[0][0] = ' ';matrix[0][n - 1] = ' ';matrix[n - 1][0] = ' ';matrix[n - 1][n - 1] = ' ';}for (int i = 0; i < n; i++) {//输出for (int j = 0; j < n; j++) {printf("%c", matrix[i][j]);}printf("\n");}}return 0;
}

3.Hello World for U（浙江大学）


评测系统

#include<iostream>
#include<string.h>
using namespace std;
int main() {string n;while (cin >> n) {int len = n.length();int n1, n2;int flag = 0;for (n1 = len - 1; n1 >= 2; n1--) {for (n2 = len - 1; n2 >= 3; n2--) {if (2 * n1 + n2 - 2 == len && n1 <= n2) {flag = 1;break;}}if (flag == 1) {break;}}flag = 0;char a[1000][1000];int num = 0;int i, j;for (i = 0; i < n1; i++) {a[i][0] = n[num];num++;}i -= 1;for (j = 1; j < n2; j++) {a[i][j] = n[num];num++;}j -= 1;for (int k = i - 1; k >= 0; k--) {a[k][j] = n[num];num++;}for (i = 0; i < n1 - 1; i++) {int ko = n2 - 2;cout << a[i][0];while (ko--) {cout << " ";}cout << a[i][n2 - 1] << endl;}for (j = 0; j < n2; j++) {cout << a[i][j];}cout << endl;}
}

二.日期问题

【求天数】
1.今年的第几天？（清华大学）

评测系统

#include<iostream>
#include<string.h>
using namespace std;
int count(int m,int y) {int sum=0;for (int i = 1; i <= m; i++) {if (i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10 || i == 12) {sum += 31;}else if (i == 2) {if ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0) {sum += 29;}elsesum += 28;}elsesum += 30;}return sum;
}
int main() {int year, month, day;while (cin >> year >> month >> day) {cout << count(month - 1, year) + day << endl;}
}

另一种方式

#include<iostream>
#include<string.h>
using namespace std;
int daytab[2][13] = {{0,31,28,31,30,31,30,31,31,30,31,30,31},//不是闰年{0,31,29,31,30,31,30,31,31,30,31,30,31},//是闰年
};
bool IsLeapYear(int year) {//是否闰年return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
}
int main() {int year, month, day;int number=0;while (scanf("%d%d%d", &year, &month,&day) != EOF) {int number = 0;int row = IsLeapYear(year);for (int j = 0; j < month; ++j) {number += daytab[row][j];}number += day;printf("%d\n", number);}return 0;
}

【求日期B】
2.打印日期（华中科技大学）

评测系统

#include<iostream>
#include<string.h>
using namespace std;
int a[2];
void count(int y,int x) {int sum=0;int premonthday2 = 0;//上个月int premonthday3 = 0;//上上个月for (int i = 1; i <= 12; i++) {if (i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10 || i == 12) {sum += 31;premonthday2 = 31;}else if (i == 2) {if ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0) {sum += 29;premonthday2 = 29;}else {sum += 28;premonthday2 = 28;}}else {sum += 30;premonthday2 = 30;}if (x < sum) {if (i != 1) {a[0] = i;}else {a[0] = 1;}a[1] = x+premonthday2-sum;if (a[1] == 0) {a[0] -= 1;a[1] = premonthday3;}return;}premonthday3 = premonthday2;}
}
int main() {int year, month, day;int daycount;while (cin >> year>> daycount) {count(year,daycount);printf("%04d-%02d-%02d\n", year, a[0], a[1]);}
}

另一种方式

#include<iostream>
#include<string.h>
using namespace std;
int daytab[2][13] = {{0,31,28,31,30,31,30,31,31,30,31,30,31},//不是闰年{0,31,29,31,30,31,30,31,31,30,31,30,31},//是闰年
};
bool IsLeapYear(int year) {//是否闰年return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
}
int main() {int year, month, day;int number;while (scanf("%d%d", &year, &number) != EOF) {month = 0;int row = IsLeapYear(year);while (number > daytab[row][month]) {number -= daytab[row][month];month++;}day = number;printf("%04d-%02d-%02d\n", year, month, day);}return 0;
}

3.日期累加（北京理工大学）

评测系统

#include<iostream>
#include<string.h>
using namespace std;
int daytab[2][13] = {{0,31,28,31,30,31,30,31,31,30,31,30,31},//不是闰年{0,31,29,31,30,31,30,31,31,30,31,30,31},//是闰年
};
bool IsLeapYear(int year) {//是否闰年return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
}
int NumberOfYear(int year) {//返回该年天数if (IsLeapYear(year)) {return 366;}elsereturn 365;
}
int main() {int m = 0;cin >> m;int year, month, day,number;while (m--) {cin >> year >> month >> day >> number;int row = IsLeapYear(year);for (int j = 0; j < month; j++) {//加上今年已经过的天数number += daytab[row][j];}number += day;while (number > NumberOfYear(year)) {//确定年number -= NumberOfYear(year);year++;}month = 0;row = IsLeapYear(year);while (number > daytab[row][month]) {//确定月number -= daytab[row][month];month++;}day = number;//确定日printf("%04d-%02d-%02d\n", year, month, day);}return 0;
}

4.日期差值（上海交通大学）

评测系统

#include<iostream>
#include<string.h>
using namespace std;
int daytab[2][13] = {{0,31,28,31,30,31,30,31,31,30,31,30,31},//不是闰年{0,31,29,31,30,31,30,31,31,30,31,30,31},//是闰年
};
bool IsLeapYear(int year) {//是否闰年return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
}
int NumberOfYear(int year) {//返回该年天数if (IsLeapYear(year)) {return 366;}elsereturn 365;
}
int main() {string s1, s2;while (cin >> s1 >> s2) {int year1, month1, day1;int year2, month2, day2;int sum = 0;year1 = (s1[0] - 48) * 1000 + (s1[1] - 48) * 100 + (s1[2] - 48) * 10 + s1[3] - 48;year2 = (s2[0] - 48) * 1000 + (s2[1] - 48) * 100 + (s2[2] - 48) * 10 + s2[3] - 48;month1 = (s1[4] - 48) * 10 + s1[5] - 48;month2 = (s2[4] - 48) * 10 + s2[5] - 48;day1 = (s1[6] - 48) * 10 + s1[7] - 48;day2 = (s2[6] - 48) * 10 + s2[7] - 48;//处理中间年份for (int i = year1 + 1; i < year2; i++) {sum += NumberOfYear(i);}//处理两端年份if (month1 == month2) {if (day1 == day2) {sum += 0;}else {sum += abs(day1 - day2)+1;}}else {int row = IsLeapYear(year1);sum += daytab[row][month1] - day1 + 1;sum += day2;if (year1 == year2) {//没跨年，月份只会增大for (int i = month1 + 1; i < month2; i++) {sum += daytab[row][i];}}else {//跨年，分两段算for (int i = month1 + 1; i <= 12; i++) {sum += daytab[row][i];}for (int i = 1; i < month2; i++) {sum += daytab[row][i];}}}cout << sum << endl;}
}

5.Day of Week（上海交通大学）

评测系统

分析：
蔡勒公式
1.基本公式
日期≤1582年10月4日↓

日期＞1582年10月4日↓

其中：
w是星期，w=0表示星期日，w=1表示星期一…
c是年份的前两位数
y是年份的后两位数
m是月份（一月m=13，二月m=14，三月m=3，四月m=4…十二月m=12）
当m=13或14时，年份要-1，即2003年变为2002年
如2002年2月5日
年份=2001
月份=14
日=5

也可采用 w=(day+2month+3(month+1)/5+year+year/4-year/100+year/400+1)%7
其中y为四位数的年份

#include<iostream>
#include<string.h>
#include<string>
using namespace std;
int main() {string s;while (getline(cin,s)) {int i;int year, year2, month, day;int yearnum, monthnum, daynum;year = month = day = year2= 0;yearnum = monthnum = daynum = 0;for (i=0; ; i++) {if (s[i] == ' ')break;day = day * 10 + s[i]-48;}daynum = i - 1;string month2;for (i += 1; ; i++) {if (s[i] == ' ')break;month2 += s[i];}if (month2 == "January") {month = 1;}else if (month2 == "February") {month = 2;}else if (month2 == "March") {month = 3;}else if (month2 == "April") {month = 4;}else if (month2 == "May") {month = 5;}else if (month2 == "June") {month = 6;}else if (month2 == "July") {month = 7;}else if (month2 == "August") {month = 8;}else if (month2 == "September") {month = 9;}else if (month2 == "October") {month = 10;}else if (month2 == "November") {month = 11;}else {month = 12;}i += 1;year = (s[i]-48) * 10 + (s[i + 1]-48);year2 = (s[i + 2]-48) * 10 + (s[i + 3]-48);if (month == 1) {month = 13;if (year2 != 0)year2--;elseyear--;}		if (month == 2) {month = 14;if (year2 != 0)year2--;elseyear--;}int wee;if((year*100+year2<1582)||(year * 100 + year2==1582&&month<10)|| (year * 100 + year2 == 1582 && month == 10&&day<=4))wee = year2 + year2 / 4 + year / 4 - 2 * year + 26 * (month + 1) / 10 + day + 2;elsewee = year2 + year2 / 4 + year / 4 - 2 * year + 26 * (month + 1) / 10 + day - 1;if (wee < 0) {wee = (wee % 7 + 7) % 7;}elsewee = wee % 7;if (wee == 0) {cout << "Sunday" << endl;}else if (wee == 1) {cout << "Monday" << endl;}else if (wee == 2) {cout << "Tuesday" << endl;}else if (wee == 3) {cout << "Wednesday" << endl;}else if (wee == 4) {cout << "Thursday" << endl;}else if (wee == 5) {cout << "Friday" << endl;}else {cout << "Saturday" << endl;}}return 0;
}

6.日期类（北京理工大学）

评测系统

#include<iostream>
#include<string.h>
#include<string>
using namespace std;
int main() {int a[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };int m = 0;cin >> m;while (m--) {int year, month, day;cin >> year >> month >> day;if (month == 12 && day == 31) {year++;day = 1;month = 1;}else if (day == a[month]) {month++;day = 1;}else {day++;}printf("%04d-%02d-%02d\n", year, month, day);}
}

三.其他模拟

1.剩下的树（清华大学）

测评系统

分析：
设置一个数组，在给定区间内就将数值设为1，最后统计1的个数

#include<iostream>
#include<string.h>
using namespace std;
int main() {int a[10005] = { 0 };int len=0, group = 0;cin >> len >> group;for (int k = 0; k < group; k++) {int m, n;cin >> m >> n;for (int i = m; i <= n; i++) {a[i] = 1;}}int sum = 0;for (int k = 0; k < len; k++) {if (a[k] == 1)sum++;}cout << len + 1 - sum;
}

2.手机键盘（清华大学）

评测系统

分析：

每一个字母需要的按键次数

int keytab[26] = { 1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,4,1,2,3,1,2,3,4 };

判断两个字母是否位于同一个按键上

若在同一按键上，两个字母的差值=按键次数的差值
例如B-A=2-1=1
C-A=3-1=2

if (str[i] - str[i - 1] == keytab[str[i] - 'a'] - keytab[str[i - 1] - 'a']) {time += 2;
}

若不在同一按键上，两个字母的差值≠按键次数的差值
G-F≠1-3

完整代码

#include<iostream>
#include<string.h>
using namespace std;
int keytab[26] = { 1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,4,1,2,3,1,2,3,4 };
int main() {string str;while (cin >> str) {int time = 0;for (int i = 0; i < str.size(); i++) {time += keytab[str[i] - 'a'];//按键时间if (i != 0 && str[i] - str[i - 1] == keytab[str[i] - 'a'] - keytab[str[i - 1] - 'a']) {time += 2;}}cout << time << endl;}return 0;
}

3.XXX定律（浙江大学）

评测系统

#include<iostream>
using namespace std;
int main() {int n = 0;while (cin>>n) {if (n == 0)break;int num = 0;while (n != 1) {if (n % 2 == 0) {n /= 2;}else {n = (3 * n + 1) / 2;}num++;}cout << num << endl;}return 0;
}

4.Grading（浙江大学）

评测系统

#include<iostream>
using namespace std;
float max(float a, float b, float c) {if (a > b) {if (a > c) return a;elsereturn c;}else {if (b > c)return b;elsereturn c;}
}
int main() {float P, T, G1, G2, G3, GJ;cin >> P >> T >> G1 >> G2 >> G3 >> GJ;if (abs(G2 - G1) <= T) {printf("%.1f", (G1 + G2) / 2);}else {if ((abs(G3 - G1) <= T && abs(G3 - G2) > T)) {printf("%.1f", (G3 + G1) / 2);}else if (abs(G3 - G1) > T && abs(G3 - G2) <= T) {printf("%.1f", (G3 + G2) / 2);}else if (abs(G3 - G1) <= T && abs(G3 - G2) <= T) {printf("%.1f", max(G1, G2, G3));}else {printf("%.1f", GJ);}}
}