qq_41427793的博客

题目描述：给一个字符串比如ABC 再给一个整数比如3.输出AAABBBCCC就行了#include <string>#include <iostream>#include <cstdio>using namespace std; int main(){ string str; getline(cin, str); int n; cin >> n; for(int i = 0; i < str.

问题描述：Mr. B, Mr. G, Mr. M and their coach Professor S are planning their way for the ACM-ICPC World Finals. Each of the four has a square-shaped suitcase with side length Ai (1<=i<=4) respectively. They want to pack their suitcases into a large squa.

数值类型一定要声明为double型的，否则上面的样例会输出YES#include<iostream>using namespace std;int main() { double a, b, c; cin >> a >> b >> c; if (a + b == c || a - b == c || b - a == c || a * b == c || a / b == c || b / a == c) cout << ...

#include<iostream>#include<algorithm>using namespace std;int main() { int x, y; int ans = 0; cin >> x >> y; while (x != y) { ans++; if (x > y) swap(x, y); y -= x; } ans++; cout << ans; system("pause")...

求线段交点，输入两组线段端点（整型），求其交点，不相交和无穷交点输出一句话就行，输出交点带小数的。输入： 0 0 5 50 2 2 0输出：1 1#include<iostream>using namespace std;int main() { int x0, y0, x1, y1; int x2, y2, x3, y3; cin >> x0 >> y0 >> x1 >> y1; cin >> x.

题目：给定一个9位数字的ISBN，求其校验位。ISBN格式为2-02-033598，校验位的计算方法如下：从左到右依次将各位数字乘10，9，8，……，2，求出其和S，作模运算得M=S mod 11。若11-M在1和9之间，校验位即为该数字；若11-M等于10，校验位为X；11-M等于11，校验位为0。输出添加校验位的ISBN，如2-02-033598-0。输入：2-02-033598输出2-02-033598-0#include<iostream>#include&

题目：给定一个整数序列，求中位数。如果序列个数为奇数，中位数为升序的中间位置，如果是偶数，这位升序的中间两个数的平均值。输入：52 1 4 3 5输出：3输入：41 4 3 2输出：2.5double型：%.lf表示四舍五入到个位%.2lf表示四舍五入到小数点后两位#include<iostream>#include<algorithm>using namespace std;const int maxn = 1..

题目：给定一个字符串，代表一个 一元一次 方程。如果有解求解，输出格式“x=数字” ，如果解的个数无穷，输出 “infinite solutions”，如果没有解输出“no solution”。字符串长度不超过 256 。输入：10x-2x-8=4x+7+x输出:x=5#include<iostream>#include<string>using namespace std;int main() { string s; cin >> s;..

题目：输入一个数n，输出前n个数的约数的和。（印象中有1s的时间限制，大数据集可能超时，比如100000000）。输入：7输出：41约数：整数A除以整数B(B≠0) 除得的商正好是整数而没有余数，我们就说A能被B整除，或B能整除A。A称为B的倍数，B称为A的约数#include<iostream>using namespace std;int gcd(int a, int b) { if (b == 0) return a; else return

#include<iostream>#include<string>using namespace std;char num[20][20] = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };char wei[20][20] = { "Shi","Bai","Qian","Wan","Yi" };int main() { string s; cin >> s; int len ...

思路：1、为什么做字符反转：每个字符串的长度难以确定，导致很难比较后缀，进行反转后，将后缀转为前缀方便比较2、使用gets一定要注意用getchar吸收多余字符3、比较完最短的字符串就可得到结果#include<iostream>#include<cstring>#include<algorithm>using namespace std;void reverse(char s[],int len) { for (int i = 0; i ..

#include<iostream>using namespace std;const int maxn = 110; //数组下标不能设太大会溢出，一般设在250000以内int main() { int a, b; int ans[maxn] = {0}; cin >> a >> b; int temp = a + b; if (temp < 0) { cout << '-'; temp = -temp; } int...

#include<iostream>#include<algorithm>using namespace std;const int maxn = 110;int main() { int a, b; int A[maxn] = { 0 }, B[maxn] = { 0 }; //char C[maxn] = { 0 }; 字符数组不初始化会输出 烫烫烫... string C; cin >> a >> b; int m = 0,...

#include<iostream>#include<string>using namespace std;const int maxn = 1100;struct stu { string name; string password;}stu,res[maxn];int main() { int n,cnt=0; bool flag; cin >> n; for (int i = 0; i < n; i++) { cin >...

#include<iostream>using namespace std;const int maxn = 10000;int main() { char mp[20][20] = { "zero","one","two","three","four","five","six","seven","eight","nine" }; int x, ans = 0; int temp[maxn] = {}; cin >> x; while (x != 0) { ...

#include<iostream>#include<string>using namespace std;int main() { string s; cin >> s; int pos = 0; while (s[pos] != 'E') //找到E的位置将底数和指数分割 pos++; int exp = 0; for (int i = pos + 2; i < s.length(); i++) //将字符型的指数化为1...

存储字符串，用二维数组#include<iostream>using namespace std;int main() { char s[90][90]; int i = 0; while (scanf("%s", s[i++]) != EOF); for (int j = i-2; j >= 0; j--){ //-2的原因是while结束i多加了一个1且字符从0开始 cout << s[j]; if(j!=0) cout<<...

1、题目有两个日期，求两个日期之间的天数输入：2013010120130105输出：42、代码#include<iostream>#include<cstdio>using namespace std;int month[13][2] = { {0,0},{31,31},{28,29},{31,31},{30,30},{31,31},{30,30},{31,31},{31,31},{30,30},{31,31},{30,30},{31,31} };

思路：1、0<A,B<10^10 则A,B都应该为long long型，输出格式为%lld2、类似还有A,B属于[-2^31,2^31]，则A+B为long long型#include<iostream>using namespace std;int main() { long long pa = 0, pb = 0; long long a, b; int da, db; cin >> a >> da >> b &gt..

1、题目：将两数之差换算为时：分：秒格式输入：123 4577973输出：12:42:592、代码#include<iostream>using namespace std;int main() { int c1, c2; int s; cin >> c1 >> c2; s = (c2 - c1) / 100; printf_s("%02d:%02d:%02d", s / 3600, s %3600/ 60, s%3600

1、题目：模拟老式手机输入，九宫格布局如下：题目输入为数字或者’-’，其中‘-’代表手机输入时等待的时间间隔，数字表示点击某个按键的次数。比如点击两次 2，则输出为 B，四次 2，则输出为 A。例：输入：255输出：AK输入（等待间隔‘-’可以无限长，也可以没有）：2222------55输出：AK2、思路...

#include<iostream>using namespace std;char mp[3] = { 'B','C','J' };int change(char c) { if (c == 'B') return 0; if (c == 'C') return 1; if (c == 'J') return 2;}int main() { char x1, x2; int n; int m1, m2; cin >> n; int times_A[3...

题目：假设有编号从 1-N 的服务器，首先给出服务器个数，再给出一组服务器状态。然后给出一次数字，表示修改状态次数，接下来输入为 i，j，x，意思是使用 x 修改从 i 到 j 的服务器。最后输出所有服务器状态例：输入：51 2 2 3 121 2 52 5 -1输出：6 6 1 2 0题意：输入一个数n，随后输入n个数，代表n个状态输入一个m随后m次操作每次输入3个数字i，j，x；将编号i-j的数加上x利用数组存储数字，for循环实现操作代码：#incl.

思路：注意数组一定要初始化（为0），否则会随机初始化成一个非常大的数#include<iostream>#include<algorithm>using namespace std;const int maxn = 11000;int main() { int A[maxn] = { 0 }, n, m, sum = 0, dis[maxn] = {0}, s, e; cin >> n; for (int i = 1; i <= n; i++..

思路：1、 数组名字不要和关键字重复，一定记得初始化2、start[i]-1的原因：考虑s[i]=13，那么花色就是13/13=1,对应为H，编号为13%13=0；实际应该是S13代码：#include<iostream>using namespace std;const int N = 54;int start[N + 1] = { 0 }, endd[N + 1] = { 0 }, nextt[N + 1] = { 0 };char mp[5] = { 'S...

1、题目：给定三个整数A,B,C，范围在[-2^63,2^63],判断A+B是否大于C2、思路：两个long long的整数相加会发生溢出，故应进行溢出判断3、代码：#include<iostream>using namespace std;int main() { int n,cnt=1; bool flag; long long a, b, c,res; cin >> n; for (int i = 0; i < n; i++).

思路：设置一个数组，数组下标为幂次，数组内容为系数代码：#include<iostream>#include<algorithm>using namespace std;const int maxn = 10010;int main() { int k1,k2,e1,e2; double x1, x2; double p[maxn] = { 0 }; cin >> k1; for (int i = 0; i < k1; i++) {.

题目：给出一个多项式，求这两个多项式的乘积代码：#include<iostream>#include<algorithm>using namespace std;const int maxn = 10010;struct poly { int e; double x;}poly[maxn];int main() { int n, m; double ans[maxn] = {0}; cin >> n; for (int i = .

思路：考试座位号要用long long来存储，超过10^10就不能用int了#include<iostream>using namespace std;const int maxn = 1100;struct stu { long long num1; int num2;}stu[maxn];int main() { int n,m,x,b,c; long long a; cin >> n; for (int i = 0; i < n; i+..

#include<iostream>#include<cstdio>using namespace std;char mp[3] = { 'W','T','L' };int main() { double x,ans=1; for (int i = 0; i < 3; i++) { double maxw = -1; int index = -1; for (int j = 0; j < 3; j++) { cin >&gt...

注意：C语言的字符串是字符数组，数组不能用另一个数组赋值，故对于char数组必须用strcpy，c++的string类型可以直接赋值#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main() { char name[5],max_name[5],min_name[5]; int year, month, day; int n,cnt=0; i...

此题做法和人口普查一样#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main() { char name[16],max_name[16],min_name[16]; int hour, minute, second; int n; int maxtime = 246060, mintime = 000000; cin >> n...

#include<iostream>using namespace std;struct stu { char name[10], gender,course[10]; int score;}stu,maxd,mind;int main() { int n,maxw=-1,minw=999999999; bool flag1, flag2; cin >> n; for (int i = 0; i < n; i++) { scanf_s("%s...

思路：代码：#include<iostream>using namespace std;int main() { int n,col; char c; cin >> n>>c; if (n % 2 == 0) col = n / 2; else col = n / 2 + 1; for (int i = 0; i < col; i++) { if (i == 0 || i == col - 1) { for (int j = ..

思路：1、先计算打印出上三角所需的行数k：由k可以推出沙漏字符总数为{[(1+2k-1)*k]/2}*2-1=2*k*k-12、打印上三角：由行数i可以推出每行空格数为i（从0开始），字符数2*(k-i)-13、打印下三角：每行空格数=（最后一行字符数-当前行字符数）/2，即[(2*k-1)-(2*i+1)]/2=k-i-1;每行字符数=2*i+1代码：#include<iostream>using namespace std;int main() { int..

#include<iostream>#include<cstring>using namespace std;int main() { char str[100]; gets_s(str); int n = strlen(str); int N = n + 2; int n1 = N / 3; int n2 = N - 2 * n1; for (int i = 0; i < n1; i++) { if (i != n1 - 1) { for ...

这道题做法和#include<iostream>#include<cmath>using namespace std;int main() { int a, b, c, d, ans = 0, res[110] = { 0 }; cin >> a >> b>>d; c = a + b; while (c != 0) { int db = c % d; c /= d; ans = ans * 10 + db; ...

错误点： 输出中间位置应该是temp %(17*29)/ 29而不是temp / 29#include<iostream>#include<cstdio>using namespace std;int main() { int a1, b1, c1, a2, b2, c2; bool flag = false; scanf("%d.%d.%d", &a1, &b1, &c1); scanf("%d.%d.%d", &a2, &a..

#include<iostream>#include<cmath>using namespace std;int main() { int a, b, res[110] = { 0 }; bool flag = false; cin >> a >> b; int i = 0; while (a != 0) { int db = a % b; a /= b; res[i++] = db; } for (int j = i -...

题目：将10进制转为13进制，其中13进制为0~9，A~C；，输出第一个字符为#代码：#include<iostream>using namespace std;int main() { char mp[13] = { '0','1','2','3','4','5','6','7','8','9','A','B','C' }; int temp; for (int i = 0; i < 3; i++) { cin >> temp; if (i ==