A - Complicated GCD
CodeForces - 664A
题意：a 到 b 的最大公约数是多少
题解：a == b 答案即为 a，否则为1
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include<stack> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long LL; const int MAXN = 2e2 + 10; const int MOD = 1e9 + 7; int main() { string a,b; cin >> a; cin >> b; if(a == b) cout << a << endl; else cout << 1 << endl; }
B - Spider Man
CodeForces - 705B
题意：给你n个数 两人游戏 若一方不能分解任何数（也就是全部被分解为1）则输
题解：计算不为 1 的数一共可以分解几次就可以了
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include<stack> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long LL; const int MAXN = 1e5 + 10; const int MOD = 1e9 + 7; int main() { int n; scanf("%d",&n); LL tmp = 0; LL x; for(int i = 0; i < n; i++) { scanf("%lld",&x); tmp += (x - 1); if(tmp == 0) puts("2"); else { if(tmp % 2 == 1) puts("1"); else puts("2"); } } }
C - Catch Overflow!
CodeForces - 1175B
题意：给出for循环伪代码，计算执行add的次数（加了几次）。
题解：两个栈，一个记录for的次数，一个记录add的数字即可
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include<stack> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long LL; const int MAXN = 1e5 + 10; const int MOD = 1e9 + 7; const LL MAXX = (1LL<<32) - 1LL; stack<LL>sta; stack<LL>stb; // 4294967296 int main() { int n; scanf("%d",&n); //cout<<MAXX<<endl; string s; LL total = 0; LL num; int flag = 0; while(n--) { cin >> s; if(s == "add") { if(sta.size() == 0) total += 1; else { LL tmp = stb.top(); stb.pop(); tmp++; //cout<<tmp<<endl; stb.push(tmp); } } else if(s == "for") { scanf("%lld",&num); if(sta.size() == 0) sta.push(num); else { //cout<<sta.top() * num<<endl; if(sta.top () * num > MAXX) sta.push(MAXX * 2); else sta.push(sta.top() * num); } stb.push(0); } else if(s == "end" && flag == 0) { LL tmp1 = sta.top(); sta.pop(); LL tmp2 = stb.top(); stb.pop(); //printf("%lld %lld ",tmp1,tmp2); total += tmp1 * tmp2; if(total > MAXX) flag = 1; } //cout<<total<<endl; } if(flag == 1 || total > MAXX) puts("OVERFLOW!!!"); else printf("%lld ",total); }
D - Maximal Area Quadrilateral
CodeForces - 340B
题意：一些点中去四个点形成四边形，求四边形的最大面积
题解：枚举两个点，寻求第三个点，用向量的乘积来计算面积
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include<stack> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long LL; const int MAXN = 3e2 + 10; const int MOD = 1e9 + 7; const LL MAXX = (1LL<<32) - 1LL; struct node{ double x,y; }p[MAXN]; double multi(node p1,node p2,node p3) { double a1,b1,a2,b2; a1 = p2.x - p1.x; a2 = p3.x - p2.x; b1 = p2.y - p1.y; b2 = p3.y - p2.y; return a1 * b2 - a2 * b1; } int main() { int n; double ans = 0; cin >> n; for(int i = 0; i < n; i++) scanf("%lf %lf",&p[i].x,&p[i].y); for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { double s1 = 0,s2 = 0; for(int k = 0; k < n; k++) { double tmp = multi(p[i],p[j],p[k]); if(tmp > 0) s1 = max(s1,tmp / 2.0); else s2 = max(s2,-tmp / 2.0); } if(s1 > 0 && s2 > 0 && s1 + s2 > ans) ans = s1 + s2; } } printf("%lf ",ans); }
E - Two Arithmetic Progressions
CodeForces - 710D
题意：x 满足 L <= x <= R,并且 x = a1k + b1 = a2l + b2,(k ,l >= 0)，求x的个数有几个
题解：找出最小解后，其余的解就是加上 a1 和 a2 的最小公倍数即可，最小解的寻找可以使用扩展中国剩余定理写，也可以直接循环逼近，这里使用的是循环逼近最小解，即一开始a1 和 a2 都为0，然后使得小的一个加上该式中的a1或a2的某倍数后使得值大于等于另一个，以此循环，如果等于了就停止，不停止的话最多逼近100000次我限制的
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include<stack> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long LL; const int MAXN = 3e2 + 10; const int MOD = 1e9 + 7; const LL MAXX = (1LL<<32) - 1LL; LL gcd(LL a,LL b) { return b == 0 ? a : gcd(b,a % b); } int main() { LL a1,b1,a2,b2,L,R; cin >> a1 >> b1 >> a2 >> b2 >> L >> R; LL lcm = a1 * a2 / gcd(a1,a2) ; //cout<<lcm<<endl; LL tmp1 = b1,tmp2 = b2; int pos = 1e5; while(tmp1 != tmp2) { if (tmp1 < tmp2) { LL tmp = tmp2 - tmp1; tmp1 += ceil(tmp * 1.0 / (a1 * 1.0)) * a1; } else { LL tmp = tmp1 - tmp2; tmp2 += ceil(tmp * 1.0 / (1.0 * a2)) * a2; } pos--; if(pos == 0) break; //printf("%lld %lld ", tmp1, tmp2); } if(pos == 0) { cout << 0 << endl; return 0; } L = max(L,max(b1,b2)); if(L > R) { cout<<0<<endl; return 0; } LL ans = 0; if(tmp1 <= R) ans += (R - tmp1) / lcm + 1; if(tmp1 < L) ans -= (L - tmp1 - 1) / lcm + 1; cout << ans << endl; }
F - Lucky Permutation
CodeForces - 286A
题意：给了一个pi数组，ppi = n - i + 1，要求求出这个数组
题解：判断-1的情况，推一推就好了，之后两两交换，再头尾交换即可
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include<stack> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long LL; const int MAXN = 1e5 + 10; const int MOD = 1e9 + 7; const LL MAXX = (1LL<<32) - 1LL; int a[MAXN]; int main() { int n; while(~scanf("%d",&n)) { if (n == 1) cout << 1 << endl; else if ((n / 2) % 2 == 1) cout << -1 << endl; else { for (int i = 1; i <= n; i++) a[i] = i; for (int i = 1; i <= n; i += 2) { if (n % 2 == 1 && i == (n + 1) / 2) i++; swap(a[i], a[i + 1]); } for (int i = 2; i <= n / 2; i += 2) swap(a[i], a[1 + n - i]); for (int i = 1; i <= n; i++) printf("%d%c", a[i], i == n ? ' ' : ' '); } } }