cf_298
来源：互联网 发布：waves gtr for mac 编辑：程序博客网 时间：2024/06/06 05:01
源链接：http://codeforces.com/contest/534
。。最近不知道为什么，，觉得做题目毫无状态。。略微迷茫。。
A：连A都错了。。真的是。。
题意就是把1-n个数排在一起，是的相邻的数相差不能为1.其实就是把奇数放在一起，，然后再把偶数放在一起。。可是事实是不能先放奇数，，比如n=4,如果先放奇数，那么就会有1 3 2 4这种情况，3和2就是差1了。。。。所以一定得先放偶数，，这样肯定不会错，，因为奇数是从1开始的，只要我们保证n=2的时候特判下，那就没问题了。。
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-5#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#define ll __int64#define pi acos(-1.0);int n,m;#define Mod 1000000007#define N 510#define M 1000100const int size = 10010;const int mod = 9901;int a[M];int mmax;int num[1010];int ans;bool cmp(int x,int y){ return x>y;}int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif int t; while(sf(n)!=EOF){ vector<int> v; if(n == 2){ printf("1\n"); printf("1\n"); continue; } if(n == 3){ printf("2\n"); printf("1 3\n"); continue; } for(int i=2;i<=n;i+=2) v.push_back(i); for(int i=1;i<=n;i+=2) v.push_back(i); int size = (int)v.size(); printf("%d\n",size); for(int i=0;i<size;i++) printf("%d%c",v[i],i+1==size?'\n':' '); }return 0;}B:这题也墨迹好久。。真的不知道在干嘛。。
其实思路不难，因为我们要保证最后一个能经过中间的过渡到达，所以除去第一个和最后一个，对于从第二个数开始，我们就计划下，如果要使得数尽量大，那么我们就取v1+i*d,但是同时要在保证v2最后能到达，所以此时的最大值是v2+(t-i)*d，要满足题意，那么我们要在这两个数中取一个较小的，即res = Min(v1+i*d,v2+(t-i)*d)，这样考虑就行。
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-5#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#define ll __int64#define pi acos(-1.0);int n,m;#define Mod 1000000007#define N 510#define M 1000100const int size = 10010;const int mod = 9901;int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif int v1,v2; int t,d; while(sfd(v1,v2)!=EOF){ scanf("%d%d",&t,&d); int ans = v1+v2; if(v1>v2) swap(v1,v2); for(int i=2;i<t;i++){ int l = v1+(i-1)*d; int r = v2+(t-i)*d; ans += Min(l,r); } printf("%d\n",ans); }return 0;}C:C题还好，刚开始考虑的比较复杂了点，其实还是挺简单的，对于每个骰子，我们能求出它能够取的最小值和最大值，那么它的范围减去它能够取的值，就是不能够取的值了。所以我们考虑下最小值和最大值，其中最大值就是当其他骰子取1的时候，此时该骰子有最大值，为A-(n-1),当然这个值要小于它本身的最大点数；最小值的话就是当其他所有骰子为最大值时，此时它能够取的值，即A-（sum-a[i])，sum就是所有骰子的总点数，当然，这最小值要大于等于1.这样子的范围求出来后，那么求不能取的范围只要总范围减去能取的范围就行了。
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-5#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64#define pi acos(-1.0);ll n,m;#define Mod 1000000007#define N 510#define M 1000100const int size = 10010;const int mod = 9901;ll a[M];int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif ll A; while(scanf("%I64d%I64d",&n,&A)!=EOF){ ll sum=0; for(int i=1;i<=n;i++){ sfI(a[i]); sum += a[i]; } if(n == 1){ printf("%I64d\n",a[1]-1); continue; } for(int i=1;i<=n;i++){ ll r = Min(A-(n-1),a[i]); ll l = Max(A-(sum-a[i]),1ll); ll res = a[i]-(r-l+1); printf("%I64d%c",res,i==n?'\n':' '); } }return 0;}D：D题是看了别人的代码，挺好理解的。模拟一下就行。对于每个握手的次数，它都有对应的一个人，那么我们开始遍历n个人，从握手次数0开始，然后握手次数+1，如果到了某个握手次数，发现没人是这个次数，那么就将其中三个人组队，握手次数-3,如果到了某个时候人数为负了，那么说明该序列是无效的。
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-5#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64#define pi acos(-1.0);int n,m;#define Mod 1000000007#define N 510#define M 1000100const int size = 10010;const int mod = 9901;queue<int> q[200005];int a[M];int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif while(scanf("%d",&n)!=EOF){ for(int i=0;i<=n;i++) while(!q[i].empty()) q[i].pop(); int x; for(int i=1;i<=n;i++){ scanf("%d",&x); q[x].push(i); } int ok = 1,peo = 0; for(int i=1;i<=n;i++){ while(1){ if(peo<0){//invalid ok=0; break; } if(!q[peo].empty()){ a[i] = q[peo].front();// q[peo].pop(); peo++; break; }else{ peo -= 3;//3 people composes a team } } if(!ok) break; } if(!ok) printf("Impossible\n"); else{ printf("Possible\n"); for(int i=1;i<=n;i++) printf("%d%c",a[i],i==n?'\n':' '); } }return 0;}CF的题质量是真的好。。。 0 0
