Codeforces Round #340 (Div. 2)题解

题目链接：点击打开链接

这次的比赛比较简单， 有些坑点， 主要是考思维的严密性。。。

比赛时做出了前4道， 都没有什么算法， 最后一道的通用解法是莫队算法， 现在还不会， 今天就补上。

终测挂了C，原来是漏了枚举第一个半径为0的情况， 太失误了。

A. Elephant

水题。

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1000000000 + 7;const ll INF = 1000000000000000000+10000000;const int maxn = 100;int T,n,m,x;ll l,r,k;int main() {    scanf("%d",&x);    int ans = x / 5;    if(ans * 5 != x) ans++;    printf("%d\n",ans);    return 0;}

B. Chocolate

注意全0的时候答案是0

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 1000000000;const int maxn = 100 + 10;int T,n,m,a[maxn];ll b[maxn];int main() {    scanf("%d",&n);    bool ok = false;    for(int i=0;i<n;i++) {        scanf("%d",&a[i]);        if(a[i]) ok = true;    }    if(!ok) { printf("0\n"); return 0; }    int len = 0;    int l = -1, r;    for(int i=0;i<n;i++) {        if(a[i] == 1) {            if(l == -1) l = i;            else {                b[len++] = i - l;                l = i;            }            continue;        }    }    ll ans = 1;    for(int i=0;i<len;i++) {        ans *= b[i];    }    printf("%I64d\n",ans);    return 0;}

C. Watering Flowers

思路：n^2枚举， 第一层枚举第一个值r^2，也就是0和其他所有点到喷泉的距离平方。 第二层再枚举一遍所有点，把所有不在第一个半径范围的点归给第二个喷泉， 然后每次更新最小值就行了。

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 1000000000;const int maxn = 2000 + 10;int T,n,m;ll x,y,x3,y3;struct node {    ll x, y;}a[maxn];int main() {    scanf("%d%I64d%I64d%I64d%I64d",&n,&x,&y,&x3,&y3);    ll max1 = 0, max2 = 0;    for(int i=0;i<n;i++) {        scanf("%I64d%I64d",&a[i].x,&a[i].y);        ll v = (x - a[i].x)*(x - a[i].x) + (y - a[i].y)*(y - a[i].y);        max1 = max(max1, v);        v = (x3 - a[i].x)*(x3 - a[i].x) + (y3 - a[i].y)*(y3 - a[i].y);        max2 = max(max2, v);    }    ll ans = max2 + max1;    for(int i=0;i<=n;i++) {        ll v = (x - a[i].x)*(x - a[i].x) + (y - a[i].y)*(y - a[i].y);        if(i == n) v = 0;        ll hehe = 0;        for(int j=0;j<n;j++) {            ll cur = (x - a[j].x)*(x - a[j].x) + (y - a[j].y)*(y - a[j].y);            if(cur <= v) continue;            cur = (x3 - a[j].x)*(x3 - a[j].x) + (y3 - a[j].y)*(y3 - a[j].y);            hehe = max(hehe, cur);        }        ll u = v + hehe;        if(ans == -1) ans = u;        else ans = min(ans, u);    }    printf("%I64d\n",ans);    return 0;}

D. Polyline

思路：不难想到， 最多三条折线就可以穿过任意3个点， 分情况讨论即可：

当三个点的x或者y坐标相同时输出1

当有两个点的x相同时， 如果第三个点的y坐标不介于另外两个点之间时，输出2； 同理当两个点y相同时......

其他情况输出3

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 1000000000;const int maxn = 100;int T,n,m;struct node {    ll x, y;}a[10];int main() {    for(int i=1;i<=3;i++) {        scanf("%I64d%I64d",&a[i].x,&a[i].y);    }    if((a[1].x == a[2].x && a[2].x == a[3].x)||(a[1].y == a[2].y &&a[2].y == a[3].y)) {        printf("1\n"); return 0;    }    else {        for(int i=1;i<=3;i++) {            for(int j=1;j<=3;j++) {                if(i == j) continue;                if(a[i].x == a[j].x) {                    int k;                    for(k=1;k<=3;k++) {                        if(k != i && k != j) break;                    }                    ll l = min(a[i].y, a[j].y);                    ll r = max(a[i].y, a[j].y);                    if(a[k].y <= l || a[k].y >= r) { printf("2\n"); return 0; }                }                if(a[i].y == a[j].y) {                    int k;                    for(k=1;k<=3;k++) {                        if(k != i && k != j) break;                    }                    ll l = min(a[i].x, a[j].x);                    ll r = max(a[i].x, a[j].x);                    if(a[k].x <= l || a[k].x >= r) { printf("2\n"); return 0; }                }            }        }        printf("3\n");    }    return 0;}

E. XOR and Favorite Number

思路：莫队算法，尽快学会，然后补上。