Codeforces Wunder fund 618ABCDEF
Codeforces Wunder fund
题目链接：http://codeforces.com/contest/618
最后一题过了12个人就不改了……
通过数 3
Standing： 506/6671
Rating change： 1811 - 1862
ABC忘记什么题了。
A：
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
int a[60];
int main()
{
int n;
while(scanf("%d", &n) != EOF){
int now = 0;
int cnt = 0;
int num = 0;
while((1 << now) <= n){
if((1 << now) & n) a[cnt++] = num + 1;
num++;
now++;
}
for(int i = cnt - 1 ; i >= 0 ; i--){
printf("%d", a[i]);
if(i == 0) printf("\n");
else printf(" ");
}
}
return 0;
}
B：
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
const int MAXN = 50 + 5;
int g[MAXN][MAXN];
int use[MAXN], ans[MAXN];
int cnt[MAXN];
int main()
{
int n;
while(scanf("%d", &n) != EOF){
memset(ans, 0, sizeof(ans));
memset(use, 0, sizeof(use));
for(int i = 1 ; i <= n ; i++) for(int j = 1 ; j <= n ; j++) scanf("%d", &g[i][j]);
for(int i = 1 ; i <= n ; i++){
for(int j = 1 ; j <= n ; j++) cnt[j] = 0;
for(int j = 1 ; j <= n ; j++) cnt[g[i][j]]++;
for(int j = 1 ; j <= n ; j++){
if(cnt[j] > 1){
ans[i] = j;
use[j] = 1;
break;
}
}
}
int head = 1;
for(int i = 1 ; i <= n ; i++){
if(ans[i] == 0){
while(use[head] == 1) head++;
ans[i] = head;
use[head] = 1;
}
}
for(int i = 1 ; i <= n ; i++){
printf("%d", ans[i]);
if(i == n) printf("\n");
else printf(" ");
}
}
return 0;
}
C：
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;
const int MAXN = 100000 + 5;
struct D
{
int x, y;
int id;
}d[MAXN];
bool cmp(D a, D b)
{
if(a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
bool cal(int u1, int v1, int u2, int v2, int u3, int v3)
{
long long t1 = v1 - v2;
t1 = t1 * (u1 - u3);
long long t2 = v1 - v3;
t2 = t2 * (u1 - u2);
// printf("t1 = %I64d, t2 = %I64d\n", t1, t2);
if(t1 == t2) return true;
else return false;
}
int main()
{
int n;
while(scanf("%d", &n) != EOF){
for(int i = 1 ; i <= n ; i++){
scanf("%d%d", &d[i].x, &d[i].y);
d[i].id = i;
}
sort(d + 1, d + 1 + n, cmp);
// for(int i = 1 ; i <= n ; i++){
// printf("x = %d, y = %d, id = %d\n", d[i].x, d[i].y, d[i].id);
// }
int rx1 = d[1].x, ry1 = d[1].y;
int rx2, ry2;
int f = 1;
int a1 = d[1].id, a2, a3;
for(int i = 2 ; i <= n ; i++){
if(f){
rx2 = d[i].x, ry2 = d[i].y;
f = 0;
a2 = d[i].id;
}
else{
if(cal(rx1, ry1, rx2, ry2, d[i].x, d[i].y)) continue;
else{
a3 = d[i].id;
break;
}
}
}
printf("%d %d %d\n", a1, a2, a3);
}
return 0;
}D：
给一个完全图，图中边有权值，其中给出的一颗生成树上的边权值为x，非此树上的边权值为y。
求不重复的走完所有点的最小代价。
赛中猜测是分类讨论。
X>y的时候不知道怎么考虑，随便写了一发。
X
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
#include <queue>
using namespace std;
#define LL long long
const int MAXN = 200000 + 5;
vector<int>lin[MAXN];
queue<int>que;
int du[MAXN];
int vis[MAXN];
int n;
LL x, y;
int dfs(int u, int pre, int &cnt)
{
int temp = 0;
for(int i = 0 ; i < (int)lin[u].size() ; i++){
int v = lin[u][i];
if(v == pre) continue;
if(dfs(v, u, cnt)) temp++;
}
if(u == 1){
cnt += max(temp - 2, 0);
return 0;
}
else{
if(temp > 1){
cnt += temp - 1;
return 0;
}
else return 1;
}
}
LL solve1()
{
for(int i = 1 ; i <= n ; i++){
if(du[i] == 1){
que.push(i);
vis[i] = 1;
break;
}
}
int cnt = 0;
dfs(1, -1, cnt);
// printf("cnt = %d\n", cnt);
return x * (n - 1 - cnt) + y * cnt;
}
LL solve2()
{
int cnt = 0;
for(int i = 1 ; i <= n ; i++){
if(du[i] == n - 1){
cnt++;
break;
}
}
return y * (n - 1 - cnt) + x * cnt;
}
int main()
{
while(scanf("%d%I64d%I64d", &n, &x, &y) != EOF){
int u, v;
for(int i = 1 ; i <= n ; i++) lin[i].clear() ,vis[i] = du[i] = 0;
while(!que.empty()) que.pop();
for(int i = 0 ; i < n - 1 ; i++){
scanf("%d%d", &u, &v);
lin[u].push_back(v);
lin[v].push_back(u);
du[u]++;
du[v]++;
}
for(int i = 1 ; i <= n ; i++) sort(lin[i].begin(), lin[i].end());
LL ans = 0;
if(x == y){
ans = x * (n - 1);
}
else if(x < y){
ans = solve1();
}
else ans = solve2();
printf("%I64d\n", ans);
}
return 0;
}E：
赛中听老耿说是计算几何没看，实际上是线段树。
线段树我也不会啊……
先导题POJ 2991，然后这题就很简单了。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;
const int MAXN = 300000 + 5;
const double PI = acos(-1.0);
int n, m;
double x[MAXN * 4], y[MAXN * 4];
int angle[MAXN * 4], flag[MAXN * 4];
void func(int o, int ang)
{
double tx = x[o], ty = y[o];
x[o] = ty * sin(ang * PI / 180.0) + tx * cos(ang * PI / 180.0);
y[o] = ty * cos(ang * PI / 180.0) - tx * sin(ang * PI / 180.0);
flag[o] = (flag[o] + ang) % 360;
angle[o] = (angle[o] + ang) % 360;
}
void push_up(int o)
{
x[o] = x[o << 1] + x[o << 1 | 1];
y[o] = y[o << 1] + y[o << 1 | 1];
}
void push_down(int o)
{
if(flag[o]){
func(o << 1, flag[o]);
func(o << 1 | 1, flag[o]);
flag[o] = 0;
}
}
void build(int l, int r, int o)
{
x[o] = y[o] = 0;
angle[o] = 90;
flag[o] = 0;
if(l == r){
x[o] = 1;
}
else{
int mid = (l + r) >> 1;
build(l, mid, o << 1);
build(mid + 1, r, o << 1 | 1);
push_up(o);
}
}
void update1(int u, int v, int L, int R, int o)
{
if(L == R){
x[o] = x[o] + v * sin(angle[o] * PI / 180.0);
y[o] = y[o] + v * cos(angle[o] * PI / 180.0);
}
else{
push_down(o);
int mid = (L + R) >> 1;
if(u <= mid) update1(u, v, L, mid, o << 1);
else update1(u, v, mid + 1, R, o << 1 | 1);
push_up(o);
}
}
void update2(int l, int r, int v, int L, int R, int o)
{
if(l <= L && r >= R){
func(o, v);
}
else{
push_down(o);
int mid = (L + R) >> 1;
if(l <= mid) update2(l, r, v, L, mid, o << 1);
if(mid < r) update2(l, r, v, mid + 1, R, o << 1 | 1);
push_up(o);
}
}
int main()
{
// freopen("CF 618E.in", "r", stdin);
while(scanf("%d%d", &n, &m) != EOF){
build(1, n, 1);
// printf("x[1] = %f, y[1] = %f\n", x[1], y[1]);
for(int i = 1 ; i <= m ; i++){
int op, u, v;
scanf("%d%d%d", &op, &u, &v);
if(op == 1){
update1(u, v, 1, n, 1);
}
else{
update2(u, n, v, 1, n, 1);
}
printf("%.10f %.10f\n", x[1], y[1]);
}
}
return 0;
}F：
两个序列，任取数使得两个序列取出的数和相等。
感觉是很重要的模型。
/*
容斥原理中的鸽笼原理。
因为前缀和是递增的，记录值SumOfa - SumOfb，其中SumOfb为小于等于SumOfa的最大值
然后由于每个数都属于[1,n - 1]，所以这个值的范围[0,n - 1]。
因为一共有n+1中取法（包括不取任意一个数，视为SumOfa - SumOfb = 0)
故必定有两个差值相等
然后看代码就知道了
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <utility>
using namespace std;
#define LL long long
const int MAXN = 1000000 + 5;
int a[MAXN], b[MAXN];
pair<int,int> vis[MAXN];
vector<int>o1, o2;
int main()
{
int n;
while(scanf("%d", &n) != EOF){
for(int i = 1 ; i <= n ; i++) scanf("%d", &a[i]);
for(int i = 1 ; i <= n ; i++) scanf("%d", &b[i]);
for(int i = 0 ; i <= n ; i++) vis[i] = make_pair(-1, -1);
vis[0] = make_pair(0, 1);
int head = 1;
LL s1 = 0, s2 = 0;
o1.clear(), o2.clear();
for(int i = 1 ; i <= n ; i++){
s1 += a[i];
while(head <= n && s2 + b[head] <= s1) s2 += b[head++];
int mark = s1 - s2;
// printf("mark = %d, i = %d, head = %d\n")
if(vis[mark].first == -1) vis[mark] = make_pair(i, head);
else{
for(int j = vis[mark].first + 1; j <= i ; j++) o1.push_back(j);
for(int j = vis[mark].second; j <= head - 1; j++) o2.push_back(j);
// printf("vis[mark] = %d %d, i = %d, head = %d\n", vis[mark].first, vis[mark].second, i, head);
// printf("s1 = %I64d, s2 = %I64d\n", s1, s2);
break;
}
}
printf("%d\n", (int)o1.size());
for(int i = 0 ; i < (int)o1.size() ; i++){
printf("%d", o1[i]);
if(i == (int)o1.size() - 1) printf("\n");
else printf(" ");
}
printf("%d\n", (int)o2.size());
for(int i = 0 ; i < (int)o2.size() ; i++){
printf("%d", o2[i]);
if(i == (int)o2.size() - 1) printf("\n");
else printf(" ");
}
}
return 0;
}