- 数学
方法
- @ 2026-1-6 19:00:59
会发题解
14 条评论
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李辰洋 LV 9 @ 2026-1-10 8:53:26有人吗
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@ 2026-1-9 13:38:45#include <bits/stdc++.h> using namespace std; int main() { int y,m,d,bm,bd,c=0; cin>>y>>m>>d>>bm>>bd; while(m!=bm || d!=bd) { c++;
d++;
int md=31; if(m4||m6||m9||m11) md=30; if(m2) { if((y%40&&y%100!=0)||y%400==0) md=29; else md=28; } if(d>md) { d=1; m++; if(m>12) { m=1; y++; } } } cout<<c<<endl; return 0; } -
@ 2026-1-9 13:38:10#include <bits/stdc++.h> using namespace std; int main() { int n,g,s,b,maxx,minx,mx,mi,temp; cin>>n; while(n!=495) { g=n%10; s=n/10%10; b=n/100%10; maxx=max(max(g,s),b); minx=min(min(g,s),b); temp=g+s+b-maxx-minx; mx=maxx100+temp10+minx; mi=minx100+temp10+maxx; n=mx-mi; cout<<mx<<"-"<<mi<<"="<<n<<endl; } return 0; }
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@ 2026-1-9 13:37:16#include <bits/stdc++.h> using namespace std; long long gcd(long long a, long long b) { while (b != 0) { long long temp = b; b = a % b; a = temp; } return a; } int main() { long long a, b, c, d; cin >> a >> b >> c >> d; long long num = a * d + c * b; long long den = b * d; long long g = gcd(num, den); cout << num / g << " " << den / g << endl; return 0; }
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@ 2026-1-9 13:36:57#include <bits/stdc++.h> using namespace std; int main() { long long a, b; cin >> a >> b; long long divisor = 10; while (divisor <= a) { long long left = a / divisor, right = a % divisor; if (left + right == b) { cout << left << "+" << right << "=" << b << endl; return 0; } divisor *= 10; } cout << "Impossible!" << endl; return 0; }
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@ 2026-1-9 13:36:29#include <bits/stdc++.h> using namespace std; int main() { long long n; cin >> n; int mf = 1000; long long temp = n; int len = 0; while (temp > 0) { len++; temp /= 10; } for (int i = 0; i <= len - 3; i++) { long long div = pow(10, len - 3 - i); int num = (n / div) % 1000; mf = min(mf, abs(num - 753)); if (mf == 0) break; } cout << mf << endl; return 0; }
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@ 2026-1-9 13:36:12#include using namespace std;
// 使用欧几里得算法计算最大公约数 long long gcd(long long a, long long b) { while (b != 0) { long long temp = b; b = a % b; a = temp; } return a; }
int main() { long long n, m; cin >> n >> m;
// 计算最大公约数作为正方形的最大边长 long long result = gcd(n, m); cout << result << endl; return 0;}
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@ 2026-1-9 13:35:52#include using namespace std; int main() { long long m, k; cin >> m >> k; bool divisibleBy19 = (m%19==0); int temp = m; int count3=0; while (temp > 0) { int digit = temp %10; if(digit ==3) { count3++; } temp = temp / 10; } if(divisibleBy19 && count3 ==k) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } return 0; }
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@ 2026-1-9 13:35:27#include using namespace std; long long gcd(long long a, long long b) { while (b != 0) { long long temp = b; b = a % b; a = temp; } return a; }
int main() { long long a, b; cin >> a >> b;
// 计算并输出最大公约数 cout << a*b/gcd(a, b) << endl; return 0;}
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@ 2026-1-9 13:35:06#include using namespace std;
// 使用欧几里得算法计算最大公约数 long long gcd(long long a, long long b) { while (b != 0) { long long temp = b; b = a % b; a = temp; } return a; }
int main() { long long a, b; cin >> a >> b;
// 计算并输出最大公约数 cout << gcd(a, b) << endl; return 0;}
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@ 2026-1-9 13:34:46#include <bits/stdc++.h> using namespace std; int main() { long long n; cin >> n; if (n == 0) { cout << 0 << endl; return 0; } bool is = false; if (n < 0) { is = true; n = -n; } long long r = 0; while (n > 0) { r = r * 10 + n % 10; n /= 10; } if (is) { r = -r; } cout << r << endl; return 0; }
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@ 2026-1-9 13:34:31#include<bits/stdc++.h> using namespace std; int main() {
int a,b=10,s=0; cin>>a; while(a!=0) { s=a%b; a=a/10; cout<<s<<" "; } return 0; } -
@ 2026-1-6 19:02:4678x^2+91y=7891
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@ 2026-1-6 19:02:1611-6 #include <bits/stdc++.h> using namespace std; int main() { int t, n; cin >> t; while (t--) { cin >> n; int k=0; for (int i=1;i*i<=n;i++) { k++; } cout << k << endl; } return 0; }
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