CodeForces - 1180A
水：
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 1e5 + 10; void chongfu(){ } int main() { ll n; scanf("%lld",&n); ll tmp = 2 * n - 1; printf("%lld\n",tmp + (tmp - 1) * (tmp - 1) / 2); }
CodeForces - 1180B
题意：给一个长度为n的数列，对于数列中的每一个数ai都可以选择把它变为-ai-1，求所构成的最大乘积
题解：对于绝对值来说的话，正数变负数的绝对值会变大，负数变正数的绝对值会变小，所以要尽可能的把数字都调为负数，但如果有奇数个数的话，要保留一个正数确保积为正
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 1e5 + 10; int a[maxn]; int main() { int n; int ling = 0; scanf("%d",&n); int zero = 0; int maxx = -INF; for(int i=0;i<n;i++) { scanf("%d",&a[i]); if(a[i] == 0) zero ++; int tmp; if(a[i] < 0) tmp = -1 * a[i] - 1; else tmp = a[i]; if(tmp > maxx) maxx = tmp; } if(n % 2 == 1 and zero == n) { for(int i=0;i<n;i++) printf("%d ",a[i]); } else if(n % 2 == 0) { for(int i=0;i<n;i++) printf("%d ",a[i] >= 0 ? -1 * a[i] - 1 : a[i]); } else if(n % 2 == 1) { for(int i=0;i<n;i++) { int tmp = a[i] > 0 ? a[i] : -1 * a[i] - 1; if(tmp == maxx) { printf("%d ", tmp); maxx = INF; } else printf("%d ",a[i] >= 0 ? -1 * a[i] - 1 : a[i]); } } }
CodeForces - 1180C
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> #include<queue> #include<deque> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 1e5 + 10; int a[maxn]; int n,q; deque<int>que; vector<pair<int,int>> v1; vector<pair<int,int>> v2; int main() { scanf("%d %d",&n,&q); int maxx = -INF; for(int i=0;i<n;i++) { scanf("%d",&a[i]); que.push_back(a[i]); maxx = max(maxx,a[i]); } while(que[0] != maxx) { int tmp1 = que[0]; int tmp2 = que[1]; que.pop_front(); que.pop_front(); if(tmp1 < tmp2) { que.push_front(tmp2); que.push_back(tmp1); } else { que.push_back(tmp2); que.push_front(tmp1); } v1.push_back({tmp1,tmp2}); } int len1 = v1.size(); n--; int len2 = n; while(n--) { int tmp1 = que[0]; int tmp2 = que[1]; que.pop_front(); que.pop_front(); if(tmp1 < tmp2) { que.push_front(tmp2); que.push_back(tmp1); } else { que.push_back(tmp2); que.push_front(tmp1); } v2.push_back({tmp1,tmp2}); } while(q--) { ll m; scanf("%lld",&m); if(m <= len1) { printf("%d %d\n",v1[m-1].first,v1[m-1].second); } else { m -= len1; m %= len2; if(m == 0) printf("%d %d\n",v2[len2-1].first,v2[len2-1].second); else printf("%d %d\n",v2[m-1].first,v2[m-1].second); } } }
CodeForces - 1180D
题意：n * m的网格中，起点为(1,1)，每次寻找一个dx，dy，然后从当前的(x,y)跳到(x + dx, y + dy)，保证每次的dx和dy不与之前的相等。要求跳完所有的格子，求跳的顺序
题解：反复从头到尾跳，再从尾到头跳 (1,1)→(n,m)→(1,2)→(n,m−1)
#include<iostream> #include<algorithm> #include<cstdio> #include<string.h> #include<math.h> #include<queue> #include<deque> using namespace std; #define ll long long const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; const int maxn = 1e5 + 10; vector<pair<int,int>>v; int main() { int n,m; scanf("%d %d",&n,&m); for(int i=1;i<=n;i++) { for (int j = 1; j <= m; j++) { v.push_back({i, j}); } } int i=0,j = v.size()-1; int tot = 0; while(true) { printf("%d %d\n",v[i].first,v[i].second); tot++; i++; if(tot == n * m) break; printf("%d %d\n",v[j].first,v[j].second); tot++; j--; if(tot == n * m) break; } }
CodeForces - 1180E
题意：给出a 和 b 数组，a为各种食物的价格，b为一列排着队的小朋友拥有的钱，小朋友排队购买食物，每个人都买自己能买的起的最贵的食物，买不起就离开队伍。给出q次操作，操作1是修改食物的价格，操作2是修改小朋友的钱，每次操作后询问当小朋友买完之后，能买到的最贵的食物的价格是多少，没有食物了就输出-1.
原文：https://www.cnblogs.com/smallhester/p/11144359.html
