1017 A除以B (20分)
其实我觉得这题就不简单，可能是我没有刷过题目，使用正常的手算的方式来解题即可。这个位置是最关键的。把第i位或者第i-1位的余数作为被除数和除数相除。
int temp1 = temp / b;
aa[i] = temp1;//除数的结果保存到aa
temp = temp % b * 10 + (a[i + 1] - '0');#include<iostream>
#include<vector>
#include<string>
using namespace std;
int main()
{
string a;
int b = 0;
int aa[1001] = { 0 };
cin >> a >> b;
int jieguo=0;
int temp = (a[0]-'0');//保留余数，同时作为被除数的第一位
if (a.size() == 1)
{
jieguo = (a[0] - '0') / b;
temp = (a[0] - '0') % b;
cout << jieguo << " " << temp;
return 0;
}
for (int i = 0; i < a.size(); i++)
{
if (i != a.size() - 1)//当到a.size（）时会数组越界需要执行else
{
int temp1 = temp / b;
aa[i] = temp1;//除数的结果保存到aa
temp = temp % b * 10 + (a[i + 1] - '0');
}
else
{
aa[i] = temp / b;
temp = temp % b;
}
}
if (aa[0] == 0)//使用上述方法输出结果会存在第一位为0的情况所以需要区分
{
for (int i = 1; i < a.size(); i++)
cout << aa[i];
cout << " " << temp;
}
else
{
for (int i = 0; i < a.size(); i++)
cout << aa[i];
cout << " " << temp;
}
return 0;
}1018 锤子剪刀布 (20分)
如果你是测试点1没有通过说明你最后的输出获胜次数最多的手势有问题。我直接用最粗暴的方式确定手势，题目的意思是要是有相同的胜利次数就按照，bcj的顺序输出
if (b[2] >= b[0] && b[2] >= b[1])maxjia = 2;
if (b[0] > b[2] && b[0] >= b[1])maxjia = 0;
if (b[1] > b[0] && b[1] > b[2])maxjia = 1;
if (b[5] >= b[3] && b[5] >= b[4])maxyi = 5;
if (b[3] > b[5] && b[3] >= b[4])maxyi = 3;
if (b[4] > b[3] && b[4] > b[5])maxyi = 4;全部代码：
#include<iostream>
#include<vector>
#include<string>
using namespace std;
int main()
{
int sum = 0;
cin >> sum;
int a[6] = { 0 };//前三个数字表示甲赢平输的个数012，后三个表示乙的345
int b[6] = { 0 };//前三个表示甲用石头剪刀布胜利的个数012，后三个表示乙的345
for (int i = 0; i < sum; i++)//c>j>b>c
{
char jia, yi;
cin >> jia >> yi;
if (jia == yi)
{
a[1]++; a[4]++; continue;
}
if (jia == 'C')
{
if(yi=='J')
{
a[0]++; a[5]++; b[0]++; continue;
}
else
{
a[2]++; a[3]++; b[5]++; continue;
}
}
if (jia == 'J')
{
if (yi == 'C')
{
a[3]++; a[2]++; b[3]++; continue;
}
else
{
a[0]++; a[5]++; b[1]++; continue;
}
}
if (jia == 'B')
{
if (yi == 'C')
{
a[0]++; a[5]++; b[2]++; continue;
}
else
{
a[3]++; a[2]++; b[4]++; continue;
}
}
}
int maxjia=2,maxyi=5;
cout << a[0] << " " << a[1] << " " << a[2] << "\n" << a[3] << " " << a[4] << " " << a[5] << "\n";
if (b[2] >= b[0] && b[2] >= b[1])maxjia = 2;
if (b[0] > b[2] && b[0] >= b[1])maxjia = 0;
if (b[1] > b[0] && b[1] > b[2])maxjia = 1;
if (b[5] >= b[3] && b[5] >= b[4])maxyi = 5;
if (b[3] > b[5] && b[3] >= b[4])maxyi = 3;
if (b[4] > b[3] && b[4] > b[5])maxyi = 4;
if (maxjia == 0)
cout << "C" << " ";
if (maxjia == 1)
cout << "J" << " ";
if (maxjia == 2)
cout << "B" << " ";
if (maxyi == 3)
cout << "C";
if (maxyi == 4)
cout << "J";
if (maxyi == 5)
cout << "B";
return 0;
}1019 数字黑洞 (20分)
测试点5需要输入是6174的时候输出一行，否则出错即：
如果忽略前缀为0的条件则测试点234错误即：0001,0023,0456
#include<iostream>
#include<vector>
#include<string>
using namespace std;
void swap(char* a, char* b) //交换两个变量
{
char temp = *a;
*a = *b;
*b = temp;
}
string up(string a)//从小到大
{
for (int i = 0; i < 3; i++)
{
int min = i;
for (int j = i + 1; j < 4; j++)
if (a[j] < a[min])
min = j;
swap(&a[min], &a[i]);
}
return a;
}
string down(string a)//从大到小
{
for (int i = 0; i < 3; i++)
{
int min = i;
for (int j = i + 1; j < 4; j++)
if (a[j] > a[min])
min = j;
swap(&a[min], &a[i]);
}
return a;
}
int main()
{
string a;
cin >> a;
if (a.length() == 3) {//出现输入只有123位数时补充位数，否则数组越界
a = a.insert(0, "0");
}
else if (a.length() == 2) {
a = a.insert(0, "00");
}
else if (a.length() == 1) {
a = a.insert(0, "000");
}
while (true)
{
if (a[0] == a[1] && a[1] == a[2] && a[2] == a[3])
{
printf("%04d - %04d = 0000", stoi(a), stoi(a));
//cout << stoi(a) << " - " << stoi(a) << " = " << "0000";
break;
}
else
{
int max = stoi(down(a)),min=stoi(up(a));
printf("%04d - %04d = %04d\n", max,min,max-min);
//cout << max << " - " << min << " = " << max-min<<"\n";
a = to_string(max - min);
if (a.length() == 3) {//出现输入只有123位数时补充位数，否则数组越界
a = a.insert(0, "0");
}
else if (a.length() == 2) {
a = a.insert(0, "00");
}
else if (a.length() == 1) {
a = a.insert(0, "000");
}
if (a == "6174")break;
}
}
return 0;
}