This commit is contained in:
HuangHai
2025-08-30 18:35:01 +08:00
commit 1f397eca87
6258 changed files with 1487241 additions and 0 deletions

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#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false); //读入输出优化的强迫症
int a[3];
cin >> a[0] >> a[1] >> a[2];
//快排
// sort(a, a + 3, greater<int>()); //降序
sort(a, a + 3, less<int>()); //升序
//输出数组
for (int i = 0; i < 3; i++) {
cout << a[i]<<" ";
}
cout << endl;
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false); //读入输出优化的强迫症
int a, b, c;
cin >> a >> b >> c;
int d = a + b + c;
float e = d * 1.0 / 3;
//设置输出为小数点后3位
cout << fixed << setprecision(3);
cout << e << endl;
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false); //读入输出优化的强迫症
printf("%.1f\n", 8.0 / 5.0); //保留一位小数
printf("%.2f\n", 8.0 / 5.0); //保留两位小数
printf("%.1f\n", 8 / 5); //实验6
printf("%d\n", 8.0 / 5.0); //实验7
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false); //读入输出优化的强迫症
int n;
cin >> n;
//求3位数的反转
int gewei = n % 10; //个位
int shiwei = n / 10 % 10; //十位
int baiwei = n / 100; //百位
cout << gewei << shiwei << baiwei << endl;
//补0输出3位数
printf("%03d",25); //结果为025,果然神奇!
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
#define maxn 105
int a[maxn];
int main() {
int x, n = 0;
while (scanf("%d", &x) == 1) a[n++] = x; //以ctrl+d结束输入
for (int i = n - 1; i >= 1; i--) printf("%d ", a[i]);
printf("%d\n", a[0]);
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
#define maxn 1010
int a[maxn];
int main() {
int n, k, first = 1;
//初始化数组
memset(a, 0, sizeof(a));
//用户输入
scanf("%d%d", &n, &k);
for (int i = 1; i <= k; i++)
for (int j = 1; j <= n; j++)
if (j % i == 0) a[j] = !a[j];
for (int i = 1; i <= n; i++)
if (a[i]) {
if (first) first = 0; else printf(" ");
printf("%d", i);
}
printf("\n");
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
#define maxn 20
int a[maxn][maxn];
//蛇型填数
int main()
{
int n, x, y, tot = 0;
scanf("%d", &n);
memset(a, 0, sizeof(a));
tot = a[x = 0][y = n - 1] = 1;
while (tot < n * n)
{
while (x + 1 < n && !a[x + 1][y]) a[++x][y] = ++tot;
while (y - 1 >= 0 && !a[x][y - 1]) a[x][--y] = ++tot;
while (x - 1 >= 0 && !a[x - 1][y]) a[--x][y] = ++tot;
while (y + 1 < n && !a[x][y + 1]) a[x][++y] = ++tot;
}
for (x = 0; x < n; x++)
{
for (y = 0; y < n; y++) printf("%3d", a[x][y]);
printf("\n");
}
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
int main() {
int count = 0;
char s[20], buf[99];
scanf("%s", s);
for (int abc = 111; abc <= 999; abc++)
for (int de = 11; de <= 99; de++) {
int x = abc * (de % 10), y = abc * (de / 10), z = abc * de;
sprintf(buf, "%d%d%d%d%d", abc, de, x, y, z);
int ok = 1;
for (int i = 0; i < strlen(buf); i++)
if (strchr(s, buf[i]) == NULL) ok = 0;
if (ok) {
printf("<%d>\n", ++count);
printf("%5d\nX%4d\n-----\n%5d\n%4d\n-----\n%5d\n\n", abc, de, x, y, z);
}
}
printf("The number of solutions = %d\n", count);
return 0;
}