Codeforce round340 div2

Codeforces round 340 div2

A：题意：每次可以走1-5步，问你最少多少步走到x。

不说了..

#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cstdio>#include <algorithm>#include <cmath>#include <ctime>using namespace std;int Ans,n;int main() {scanf("%d",&n);Ans = n / 5 + ((n % 5 == 0) ? 0 : 1);cout<<Ans;return 0; }

B：题意：将一个0,1序列分成若干只有1的序列的方案数。

首先没有1的话答案是零（坑死）。

否则的话我们看两两1之间0的个数，设问num，那么把所有num + 1连乘起来就是答案了。

#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cstdio>#include <algorithm>#include <cmath>#include <ctime>#define Int long longusing namespace std;Int n,A[1010],head,tail,Ans = 1;bool F = false;int main() {scanf("%I64d",&n);for(int i = 1;i <= n;i ++){scanf("%I64d",&A[i]);if(A[i] == 1) F = true;}if(!F) printf("0"),exit(0);head = 1;while(head <= n){while(head <= n && A[head] == 1) head++;if(head > n) break;tail = head;while(A[tail + 1] == 0 && tail < n) tail ++;if(head == 1 || tail == n) head = tail + 1;else Ans *= (tail - head + 2),head = tail + 1;}cout<<Ans;return 0;}

C：题意：给出平面上n个点，问以两个给定点为圆心作园，覆盖这n个点所需的两园半径的最小平方和是多少。

我们把n个点按与第一个点的距离排序，记录前缀和need[i]表示覆盖第1-i所需的最小的r1，记录后缀和have[i]表示覆盖i-n所需的最小的r2，那么min(need[i] + haveh[I + 1])就是答案。

#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cstdio>#include <algorithm>#include <cmath>#include <ctime>#define Int long longusing namespace std;struct point{Int x;Int y;};point Point[100010];Int Ans = 1e18,n,a,b,c,d,have[100010],need[100010];Int calc(Int A,Int B,Int C,Int D) {return (A - C) * (A - C) + (B - D) * (B - D);}bool comp(const point &x,const point &y) {Int val1 = calc(a,b,x.x,x.y);Int val2 = calc(a,b,y.x,y.y);return val1 < val2;}int main() {cin>>n;cin>>a>>b>>c>>d;for(int i = 1;i <= n;i ++) cin>>Point[i].x>>Point[i].y;sort(Point + 1,Point + n + 1,comp);for(int i = 1;i <= n;i ++) need[i] = max(calc(a,b,Point[i].x,Point[i].y),need[i - 1]);for(int i = n;i >= 1;i --) have[i] = max(calc(c,d,Point[i].x,Point[i].y),have[i + 1]);for(int i = 0;i <= n;i ++) Ans = min(Ans,need[i] + have[i + 1]);cout<<Ans;}

D：题意：给出三个点，用一条折线经过这三个点，折线的所有部分必须和坐标轴平行，问折线最少由几条直线组成。

大特判..

若三点一线，答案是1.

若两点一线且三点不是构成T字型，答案是2

否则答案是3.

#include <cstring>#include <cstdlib>#include <string>#include <cstdio>#include <algorithm>#include <cmath>#include <ctime>using namespace std;int Ans,n;int a,b,c,d,e,f;int main() {scanf("%d%d",&a,&b);scanf("%d%d",&c,&d);scanf("%d%d",&e,&f);if((a == c && c == e) || (b == d && d == f)) printf("1");else if(a == c){if(f >= max(b,d) || f <= min(b,d)) printf("2");else printf("3");}else if(a == e){if(d >= max(b,f) || d <= min(b,f)) printf("2");else printf("3");}else if(c == e){if(b >= max(f,d) || b <= min(f,d)) printf("2");else printf("3");}else if(b == d){if(e >= max(a,c) || e <= min(a,c)) printf("2");else printf("3");}else if(b == f){if(c >= max(a,e) || c <= min(a,e)) printf("2");else printf("3");}else if(d == f){if(a >= max(e,c) || a <= min(e,c)) printf("2");else printf("3");}else printf("3");return 0;}

E：题意：给出一个序列，每次询问l到r间A[i]^A[i+1]^..A[j] ==k的i，j的数量（i<=j,k是全局的变量，不变。）

莫队算法，首先A[i]^A[i+1]..A[j]=(A[1]^A[2]..A[i-1])^(A[1]^A[2]…A[j])（i<=j）

那么我们记录前缀和s[]。问题就变成了每次询问l到r内s[i-1]^s[j]==k的对数。

于是再记录两个桶分别表示s[i-1]和s[i]出现的次数，就可以了。（注意每次加一个点是加在前面还是后面，加在前面用s[j]桶更新，否则用s[j-1]的桶）。.

#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cstdio>#include <algorithm>#include <cmath>#include <ctime>#define Int long longusing namespace std;struct Ques {Int l;Int r;Int id;Int Ans;};Ques q[100010];Int n,m,A[100010],s[100010],head,tail,v,k;Int Ans,M[3000010],N[3000010];bool comp(const Ques &x,const Ques &y) {if(x.l / v != y.l / v) return x.l < y.l;return x.r < y.r;}bool back(const Ques &x,const Ques &y) {return x.id < y.id;}int main() {scanf("%I64d%I64d%I64d",&n,&m,&k);v = (Int)sqrt(double(n));for(Int i = 1;i <= n;i ++) scanf("%I64d",&A[i]);for(Int i = 1;i <= n;i ++) s[i] = s[i - 1] ^ A[i];for(Int i = 1;i <= m;i ++){scanf("%I64d%I64d",&q[i].l,&q[i].r);q[i].id = i;}sort(q + 1,q + m + 1,comp);head = 1,tail = 0;Ans = 0;for(Int i = 1;i <= m;i ++){while(tail < q[i].r){tail ++;N[s[tail]] ++;M[s[tail - 1]] ++;Ans += M[k ^ s[tail]];}while(tail > q[i].r){Ans -= M[k ^ s[tail]];N[s[tail]] --;M[s[tail - 1]] --;tail --;}while(head < q[i].l){Ans -= N[k ^ s[head - 1]];N[s[head]] --;M[s[head - 1]] --;head ++;}while(head > q[i].l){head --;N[s[head]] ++;M[s[head - 1]] ++;Ans += N[k ^ s[head - 1]];}q[i].Ans = Ans;}sort(q + 1,q + m + 1,back);for(Int i = 1;i <= m;i ++) printf("%I64d\n",q[i].Ans);return 0;}