Codeforces Round #337 (Div. 2) 解题报告

苟蒻还是现场A了三题场外A掉D题。。。



A:

题意：给一根木棍将它分为整数的四段，使其刚好围成一个长方形，问有几种方案？？(全排列哦。。。)



奇数不算，正方形不算。。。。。233

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

int n,m,ans;

int main()
{
#ifdef YZY
freopen("yzy.txt","r",stdin);
#endif

cin >> n;
if (n % 2 == 1) {
cout << 0;
return 0;
}
m = n/4;
if (m*4 == n) cout << m-1;
else cout << m;
return 0;
}



B:

题意：Vika 有n桶油漆，每桶可以刷ki次，她想从第i种开始刷，刷一次换下一桶，问她最多能这样循环几次？？



这种水题WA一排我也是。。。。

循环次数肯定不超过min{ki}，于是先找出这个数，在刷的时候显然以某捅最小的后面一个为起点最优，那么就找呀找看哪两桶这样的之间别的油漆的数量最多bulabulabula...........

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn = 2E5 + 20;
typedef long long LL;

LL num[maxn],ans = 0,First,now,Last,Max = 0;
int n,i,j,pos,tmp = 2E9;

int main()
{
#ifdef YZY
freopen("yzy.txt","r",stdin);
#endif

cin >> n;
for (i = 1; i <= n; i++) {
scanf("%d",&num[i]);
if (num[i] < tmp) {
tmp = num[i];
}
}

First = now = 0;
for (i = 1; i <= n; i++) {
num[i] -= tmp;
if (!num[i]) {
if (!First) {
First = Last = i;
}
else {
Max = max(Max,1LL*(i-Last-1));
Last = i;
}
}
}
Max = max(Max,1LL*(n-Last+First-1));
ans = 1LL*n*1LL*tmp;
ans += Max;
cout << ans;
return 0;
}
C:

题意：构造一个哈达玛矩阵。。。。



显然分治！

找找个小块之间的规律和联系~

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn = 1<<10;

int i,j,p[maxn][maxn],n;

void solve(int k)
{
for (i = 1; i <= (1<<(k-1)); i++)
for (j = 1; j <= (1<<(k-1)); j++)
p[i][j+(1<<(k-1))] = p[i][j]^1;

for (i = 1+(1<<(k-1)); i <= (1<<k); i++)
for (j = 1; j <= (1<<(k-1)); j++)
p[i][j] = p[i][j+(1<<(k-1))] = p[i-(1<<(k-1))][j];
}

int main()
{
#ifdef YZY
freopen("yzy.txt","r",stdin);
#endif

cin >> n;
memset(p,0,sizeof(p));
p[1][1] = 1;
for (int l = 1; l <= n; l++) solve(l);
for (i = 1; i <= (1<<n); i++) {
for (j = 1; j <= (1<<n); j++)
if (p[i][j] == 1) printf("+");
else printf("*");
printf("\n");
}
return 0;
}


D:

题意：给一堆线段求面积并

离散化+线段树乱搞之~

就是扫描线从下往上扫，然后线段树每个点维护的那段是这个区间加上该区间往右的那个数和这个区间的距离。。



#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 1E5 + 10;
typedef long long LL;

LL sum[maxn*20],num[2*maxn],cover[maxn*20];

struct Li{
LL Y,L,R,w;
bool operator < (const Li &b) const {
return Y < b.Y;
}
Li() {}
Li(LL Y,LL L,LL R,LL w) : Y(Y),L(L),R(R),w(w) {}
}Line[2*maxn];

int n;

void maintain(int o,int l,int r)
{
if (cover[o]) {
sum[o] = num[r+1] - num[l];
return;
}

if (l == r) {
sum[o] = 0;
return;
}

sum[o] = sum[2*o] + sum[2*o+1];
}

void Modify(int o,int l,int r,int ql,int qr,LL w)
{
if (ql <= l && r <= qr) {
cover[o] += w;
maintain(o,l,r);
return;
}

int mid = (l+r) >> 1;
if (ql <= mid) Modify(2*o,l,mid,ql,qr,w);
if (qr > mid) Modify(2*o+1,mid+1,r,ql,qr,w);
maintain(o,l,r);
}

int main()
{
#ifdef YZY
freopen("yzy.txt","r",stdin);
#endif

cin >> n;
int cur = 0;
for (int i = 1; i <= n; i++) {
int X1,Y1,X2,Y2;
scanf("%d%d%d%d",&X1,&Y1,&X2,&Y2);
if (X1 > X2) swap(X1,X2);
if (Y1 > Y2) swap(Y1,Y2);
++X2; ++Y2;
Line[i] = Li(Y1,X1,X2,1);
Line[i+n] = Li(Y2,X1,X2,-1);
num[++cur] = 1LL*X1; num[++cur] = 1LL*X2;
}

sort(Line + 1,Line + 2*n + 1);
sort(num + 1,num + cur + 1);
cur = unique(num + 1,num + cur + 1) - num - 1;
num[cur + 1] = num[cur];

LL ans = 0;
for (int i = 1; i < 2*n; i++) {
int L = lower_bound(num + 1,num + cur + 1,Line[i].L) - num;
int R = lower_bound(num + 1,num + cur + 1,Line[i].R) - num - 1;
Modify(1,1,cur,L,R,Line[i].w);
ans += sum[1]*(Line[i+1].Y - Line[i].Y);
}
cout << ans;
return 0;
}