Solved:4
Rank:122
补题:8/10
A digits 2
签到 把这个数写n遍
#include <bits/stdc++.h>
using namespace std;
int T;
int n;
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i=1;i<=n;i++)printf("%d",n);
puts("");
}
return 0;
}
B generator 1
题意:求矩阵快速幂 幂超级大
题解:10进制模拟矩阵快速幂
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mod;
struct node {
ll c[2][2];
};
node mul(node x, node y) {
node res;
memset(res.c, 0, sizeof(res.c));
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++)
for(int k = 0; k < 2; k++)
res.c[i][j] = (res.c[i][j] + x.c[i][k] * y.c[k][j] % mod) % mod;
return res;
}
node pow_mod(node x, ll y) {
node res;
res.c[0][0] = res.c[1][1] = 1LL;
res.c[0][1] = res.c[1][0] = 0LL;
while(y) {
if(y & 1) res = mul(res, x);
x = mul(x, x);
y >>= 1;
}
return res;
}
char s[1000005];
int ss[1000005];
int main() {
ll x0, x1, a, b;
cin>>x0>>x1>>a>>b;
scanf("%s", s + 1);
int len = strlen(s + 1);
for(int i = 1; i <= len; i++) ss[i] = s[i] - '0';
cin>>mod;
node A;
A.c[0][0] = a;
A.c[0][1] = b;
A.c[1][0] = 1LL;
A.c[1][1] = 0LL;
if(len == 1) {
if(s[1] == '1') {
printf("%lld\n", x1);
return 0;
}
}
ss[len]--;
for(int i = len; i >= 1; i--) {
if(ss[i] < 0) {
ss[i] += 10;
ss[i - 1]--;
}
}
node now;
now.c[0][0] = now.c[1][1] = 1LL; now.c[0][1] = now.c[1][0] = 0LL;
for(int i = 1; i <= len; i++) {
now = pow_mod(now, 10);
int t = ss[i];
node pp = pow_mod(A, 1LL * t);
now = mul(now, pp);
}
ll ans = now.c[0][0] * x1 % mod + now.c[0][1] * x0 % mod;
ans %= mod;
printf("%lld\n", ans);
return 0;
}
C generator 2 (BSGS)
只是大概了解了下BSGS的大概思想.. 题是学弟补的 orz (我摸鱼了
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1000000;
int T;
ll x0,n,a,b,p;
ll ksm(ll a,ll b){
ll res = 1;
for(;b;b>>=1){
if(b & 1)res = res*a%p;
a = a * a % p;
}
return res;
}
pair<int,int> node[maxn];
int pos[maxn],val[maxn];
int main(){
scanf("%d",&T);
while(T--){
scanf("%lld%lld%lld%lld%lld",&n,&x0,&a,&b,&p);
int Q;scanf("%d",&Q);
if(a == 0){
while(Q--){
int v;scanf("%d",&v);
if(v == x0)
puts("0");
else if(v == b) puts("1");
else puts("-1");
}
continue;
}
ll now = x0;
int up = min(n,(ll)maxn);
for(int i=0;i<up;i++){
node[i] = {now,i};
now = (now * a + b) % p;
}
sort(node,node+up);
int m = 0;
for(int i=0;i<up;i++){
val[m] = node[i].first;
pos[m++] = node[i].second;
while(i < up - 1 && node[i].first == node[i+1].first)i++;
}
int inv_a = ksm(a,p-2);
int inv_b = (p-b)%p*inv_a%p;
ll bb = 0, aa = 1;
for(int i=0;i<maxn;i++){
aa = aa * inv_a % p;
bb = (bb*a%p + b) % p;
}
int r = p / maxn + 1;
while(Q--){
int v;scanf("%d",&v);
int id = lower_bound(val,val+m,v) - val;
if(id < m && val[id] == v){
printf("%d\n",pos[id]);continue;
}
if(n < maxn){
puts("-1");continue;
}
bool flag = false;
for(int i=1;i<=r;i++){
v = (v-bb+p)%p*aa%p;
int id = lower_bound(val,val+m,v) - val;
if(id < m && val[id] == v){
int res = pos[id] + i * maxn;
if(res >= n)puts("-1");
else printf("%d\n",res);
flag = true;break;
}
}
if(!flag)puts("-1");
}
}
return 0;
}
E independent set 1 (状压DP)
题意:最多26个点 一共有1<<26个子集 求所有子集最大独立集的和
题解:看题解补的.. 看到26个点以为int根本开不下状压的数组 结果可以用char开.. 因为每个状态的值不会超过26.. (原来还可以这样
于是对于每一个状态 选取他最小的一个点 可能有两种状态转移过来 选了这个点的贡献 和不选这个点
为什么是最小的就可以.. 因为当前状态是1个1个点慢慢选出来的 去掉这个点是子问题
还有为什么在算这个点的时候 要把和他相邻的点都扣掉.. 我最开始觉得有的点虽然和他相邻 但是不一定在那个状态下这个点产生了贡献 所以应该不扣
扣掉的点肯定不会对最大独立集产生贡献 因为是要放入的点相邻的 然后扣掉是不扣的子问题 所以没区别了
#include <bits/stdc++.h>
using namespace std;
char dp[1 << 26];
int adj[30];
int main() {
int n, m;
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; i++) {
int u, v;
scanf("%d%d", &u, &v);
adj[u] |= (1 << v);
adj[v] |= (1 << u);
}
for(int i = 0; i < n; i++) adj[i] = (~adj[i]), adj[i] ^= (1 << i);
//puts("??");
for(int i = 1; i < (1 << n); i++) {
int t = __builtin_ctz(i);
dp[i] = max(dp[i - (1 << t)], (char)(dp[i & adj[t]] + 1));
}
int ans = 0;
for(int i = 1; i < (1 << n); i++) ans += dp[i];
printf("%d\n", ans);
return 0;
}
F maximum clique 1 (最大独立集)
题意:n个数 选一个最大的子集 使得子集的元素两两不矛盾 矛盾是指两个数的二进制至少有两位不一样
题解:不矛盾的话 显然两个数xor起来不是2的幂和0 而且我们发现如果两个数的二进制1的个数奇偶性一样的话 是一定不矛盾的
于是根据二进制1的个数 建二分图 我跑的最大独立集 = 最小割
输出匹配的话 如果在最后一次bfs中 dis为INF 且是二分图中连s点的集合的点 那么就是被割掉了
dis不为INF的点 如果是连t点的集合中的点 也是被割掉了 因为这个点是肯定跑不到t点的
map常数真的大.... 学了下出题人std判2的幂
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
map<int, int> mp;
int n, s, t, maxflow, cnt;
struct node {
int to, nex, val;
}E[1300000];
int head[5005];
int cur[5005];
void addedge(int x, int y, int va) {
E[++cnt].to = y; E[cnt].nex = head[x]; head[x] = cnt; E[cnt].val = va;
E[++cnt].to = x; E[cnt].nex = head[y]; head[y] = cnt; E[cnt].val = 0;
}
int dis[5005];
bool inque[5005];
int st[1000005];
bool spfa() {
for(int i = 1; i <= t; i++) dis[i] = INF, inque[i] = 0, cur[i] = head[i];
int top = 0;
dis[s] = 0, inque[s] = 1;
st[++top] = s;
while(top) {
int u = st[top];
top--;
inque[u] = 0;
for(int i = head[u]; i; i = E[i].nex) {
int v = E[i].to;
if(E[i].val && dis[v] > dis[u] + 1) {
dis[v] = dis[u] + 1;
if(!inque[v]) {
inque[v] = 1;
st[++top] = v;
}
}
}
}
return dis[t] != INF;
}
int vis;
int dfs(int x, int flow) {
if(x == t) {
vis = 1;
maxflow += flow;
return flow;
}
int used = 0;
int rflow = 0;
for(int i = cur[x]; i; i = E[i].nex) {
cur[x] = i;
int v = E[i].to;
if(E[i].val && dis[v] == dis[x] + 1) {
if(rflow = dfs(v, min(flow - used, E[i].val))) {
used += rflow;
E[i].val -= rflow;
E[i ^ 1].val += rflow;
if(used == flow) break;
}
}
}
return used;
}
void dinic() {
maxflow = 0;
while(spfa()) {
vis = 1;
while(vis) {
vis = 0;
dfs(s, INF);
}
}
}
bool is_power2(int x) {
return x && (x & (x - 1)) == 0;
}
int a[5005];
int er[5005];
int main() {
scanf("%d", &n);
cnt = 1;
s = n + 1;
t = s + 1;
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
int tot = 0;
int tmp = a[i];
while(tmp) {
if(tmp & 1) tot++;
tmp >>= 1;
}
if(tot & 1) addedge(s, i, 1), er[i] = 1;
else addedge(i, t, 1), er[i] = 0;
}
for(int i = 1; i <= n; i++) {
for(int j = i + 1; j <= n; j++) {
if(is_power2(a[i] ^ a[j])) {
if(er[i]) addedge(i, j, 1);
else addedge(j, i, 1);
}
}
}
dinic();
int ans = n - maxflow;
printf("%d\n", ans);
int cn = 0;
for(int i = 1; i <= n; i++) {
if(dis[i] == INF) {
if(!er[i]) {
cn++;
if(cn != ans) printf("%d ", a[i]);
else {
printf("%d\n", a[i]);
break;
}
}
} else {
if(er[i]) {
cn++;
if(cn != ans) printf("%d ", a[i]);
else {
printf("%d\n", a[i]);
break;
}
}
}
}
return 0;
}
G subsequence 1 (DP)
题意:给s，t的数字串 求s中有多少子序列10进制下大于t
题解:dp i,j,k表示s的第i位匹配到t的第j位 k = 0表示这个子序列小于t的前j位 1表示等于 2表示大于 模拟即可
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 998244353;
char s[3005];
char t[3005];
ll dp[3005][3005][3];
int main() {
int T;
scanf("%d", &T);
while(T--) {
int n, m;
scanf("%d%d", &n, &m);
scanf("%s", s + 1);
scanf("%s", t + 1);
for(int i = 1; i <= n; i++) {
int t1 = s[i] - '0';
for(int j = 0; j <= i; j++) dp[i][j][0] = dp[i][j][1] = dp[i][j][2] = 0;
//dp[i][0][1] = 1;
//dp[i][0][1] = 1;
//dp[i][0][2] = 1;
for(int j = 1; j <= i; j++) {
int t2 = t[j] - '0';
dp[i][j][0] = dp[i - 1][j][0];
dp[i][j][1] = dp[i - 1][j][1];
dp[i][j][2] = dp[i - 1][j][2];
if(j == 1) {
if(t1 == 0) continue;
if(t1 == t2) dp[i][j][1]++;
else if(t1 > t2) dp[i][j][2]++;
else dp[i][j][0]++;
} else if(j == m + 1) {
dp[i][j][2] += dp[i - 1][j - 1][0] + dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2];
dp[i][j][1] = 0;
dp[i][j][0] = 0;
} else if(j > m + 1) {
dp[i][j][2] += dp[i - 1][j - 1][2];
}
else {
if(t1 == t2) {
dp[i][j][1] += dp[i - 1][j - 1][1];
dp[i][j][2] += dp[i - 1][j - 1][2];
dp[i][j][0] += dp[i - 1][j - 1][0];
} else if(t1 > t2) {
dp[i][j][2] += dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2];
dp[i][j][0] += dp[i - 1][j - 1][0];
} else {
dp[i][j][0] += dp[i - 1][j - 1][1] + dp[i - 1][j - 1][0];
dp[i][j][2] += dp[i - 1][j - 1][2];
}
}
dp[i][j][0] %= mod;
dp[i][j][1] %= mod;
dp[i][j][2] %= mod;
}
}
ll ans = 0;
/*
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= i; j++) {
printf("%d %d %lld %lld %lld\n", i, j, dp[i][j][0], dp[i][j][1], dp[i][j][2]);
}
}*/
for(int i = m; i <= n; i++) {
ans = (ans + dp[n][i][2]) % mod;
}
printf("%lld\n", ans);
}
return 0;
}
H subsequence 2 (模拟)
题意:给一个串的许多信息 每次给两种字母 然后给出在只剩下这两种字母的情况下原串的样子 输出原串
题解:每种信息更新这一种字母前有多少字母 最后组成原串
#include <bits/stdc++.h>
using namespace std;
char s[5];
char t[10005];
int num[15];
int pos[15][10005];
int ans[10005];
int main() {
int n, m;
scanf("%d%d", &n, &m);
for(int i = 1; i <= 10; i++) pos[i][0] = 0;
for(int i = 1; i <= n; i++) ans[i] = 0;
bool f = true;
for(int i = 1; i <= m * (m - 1) / 2; i++) {
scanf("%s", s + 1);
int len2; scanf("%d", &len2);
getchar();
if(len2 == 0) continue;
scanf("%s", t + 1);
int c1 = s[1] - 'a' + 1;
int c2 = s[2] - 'a' + 1;
int cn1 = 0, cn2 = 0;
if(c1 > c2) swap(c1, c2);
for(int j = 1; j <= len2; j++) {
if(t[j] - 'a' + 1 == c1) {
cn1++;
pos[c1][cn1] += cn2;
} else if(t[j] - 'a' + 1 == c2) {
cn2++;
pos[c2][cn2] += cn1;
} else {
f = false;
break;
}
}
if(pos[c1][0] == 0) pos[c1][0] = cn1;
else if(pos[c1][0] != cn1) f = false;
if(pos[c2][0] == 0) pos[c2][0] = cn2;
else if(pos[c2][0] != cn2) f = false;
}
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= pos[i][0]; j++) {
//cout << pos[i][j] << " ";
if(pos[i][j] + j > n) f = false;
else {
if(ans[pos[i][j] + j]) f = false;
else ans[pos[i][j] + j] = i;
}
}
//puts("");
}
//puts("f");
//cout << f <<endl;
for(int i = 1; i <= n; i++) if(!ans[i]) f = false;
if(!f) {
puts("-1");
} else {
for(int i = 1; i <= n; i++) {
printf("%c", ans[i] - 1 + 'a');
}
puts("");
}
return 0;
}
I three points 1 (计算几何)
队友补的 (啥东西写了400行???
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int gcd(int a, int b){
return b == 0 ? a : gcd(b, a % b);
}
const double eps = 1e-8;
int cmp(double x) {
if (fabs(x) < eps) return 0;
if (x > 0) return 1;
return -1;
}
const double pi = acos(-1.0);
inline double sqr(double x) {
return x * x;
}
struct point {
double x, y;
point() {}
point(double a, double b) : x(a), y(b) {}
void input() {
scanf("%lf%lf", &x, &y);
}
friend point operator + (const point &a, const point &b) {
return point(a.x + b.x, a.y + b.y);
}
friend point operator - (const point &a, const point &b) {
return point(a.x - b.x, a.y - b.y);
}
friend bool operator == (const point &a, const point &b) {
return cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0;
}
friend point operator * (const point &a, const double &b) {
return point(a.x * b, a.y * b);
}
friend point operator * (const double &a, const point &b) {
return point(a * b.x, a * b.y);
}
friend point operator / (const point &a, const double &b) {
return point(a.x / b, a.y / b);
}
double norm(){
return sqrt(sqr(x) + sqr(y));
}
friend bool operator < (const point a, const point b){
if(a.x == b.x) return a.y < b.y;
if(a.y == b.y) return a.x < b.x;
return a.x < b.x;
}
};
double det(const point &a, const point &b) {
return a.x * b.y - a.y * b.x;
}
double dot(const point &a, const point &b) {
return a.x * b.x + a.y * b.y;
}
double dist(const point &a, const point &b) {
return (a - b).norm();
}
point rotate_point(const point &p, double A){
double tx = p.x, ty = p.y;
return point(tx * cos(A) - ty * sin(A), tx * sin(A) + ty * cos(A));
}
struct line {
point a, b;
line() {}
line(point x, point y) : a(x), b(y) {}
};
line point_make_line(const point a, const point b) {
return line(a, b);
}
double dis_point_segment(const point p, const point s, const point t) {
if(cmp(dot(p - s, t - s)) < 0) return (p - s).norm();
if(cmp(dot(p - t, s - t)) < 0) return (p - t).norm();
return fabs(det(s - p, t - p) / dist(s, t));
}
void PointProjLine(const point p, const point s, const point t, point &cp){
double r = dot((t - s), (p - s)) / dot(t - s, t - s);
cp = s + r * (t - s);
}
bool PointOnSegment(point p, point s, point t){
return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0;
}
bool parallel(line a, line b){
return !cmp(det(a.a - a.b, b.a - b.b));
}
bool line_make_point(line a, line b, point &res){
if(parallel(a, b)) return false;
double s1 = det(a.a - b.a, b.b - b.a);
double s2 = det(a.b - b.a, b.b - b.a);
res = (s1 * a.b - s2 * a.a) / (s1 - s2);
return true;
}
line move_d(line a, const double &len){
point d = a.b - a.a;
d = d / d.norm();
d = rotate_point(d, pi / 2);
return line(a.a + d * len, a.b + d * len);
}
bool SegmentSamePoints(line i, line j){
return max(i.a.x, i.b.x) >= min(j.a.x, j.b.x) &&
max(j.a.x, j.b.x) >= min(i.a.x, i.b.x) &&
max(i.a.y, i.b.y) >= min(j.a.y, j.b.y) &&
max(j.a.y, j.b.y) >= min(i.a.y, i.b.y) &&
cmp(det(j.a - i.a, i.b - i.a) * det(j.b - i.a, i.b - i.a)) <= 0 &&
cmp(det(i.a - j.a, j.b - j.a) * det(i.b - j.a, j.b - j.a)) <= 0;
}
double Sabc(point a, point b, point c){
return fabs(a.x * b.y + b.x * c.y + c.x * a.y - a.x * c.y - b.x * a.y - c.x * b.y) / 2.0;
}
struct section{
double l, r;
section() {}
section(double a, double b) {
l = min(a, b);
r = max(a, b);
}
friend bool operator < (section a, section b){
if(a.l == b.l) return a.r < b.r;
return a.l < b.l;
}
};
bool sjs(section a, section b){
if(b < a) swap(a, b);
if(a.l < b.l){
return (a.r > b.l) || (b.r < a.r);
}
else{
return (a.r < b.r);
}
}
const int maxn = 1005;
struct polygon {
int n;
point a[maxn];
polygon() {}
double perimeter() {
double sum = 0;
a[n] = a[0];
for(int i = 0; i < n; i++){
sum += (a[i + 1] - a[i]).norm();
}
return sum;
}
double area(){
double sum = 0;
a[n] = a[0];
for(int i = 0; i < n; i++){
sum += det(a[i + 1], a[i]);
}
return sum / 2.0;
}
int Point_In(point t){
int num = 0, i, d1, d2, k;
a[n] = a[0];
for(i = 0; i < n; i++){
if(PointOnSegment(t, a[i], a[i + 1])) return 2;
k = cmp(det(a[i + 1] - a[i], t - a[i]));
d1 = cmp(a[i].y - t.y);
d2 = cmp(a[i + 1].y - t.y);
if(k > 0 && d1 <= 0 && d2 > 0) num++;
if(k < 0 && d2 <= 0 && d1 > 0) num--;
}
return num != 0;
}
int Border_Int_Point_Num();
int Inside_Int_Point_Num();
};
int polygon::Border_Int_Point_Num(){
int num = 0;
a[n] = a[0];
for(int i = 0; i < n; i++){
num += gcd(abs(int(a[i + 1].x - a[i].x)), abs(int(a[i + 1].y - a[i].y)));
}
return num;
}
int polygon::Inside_Int_Point_Num(){
return int(area()) + 1 - Border_Int_Point_Num() / 2;
}
struct polygon_convex{
vector<point> P;
polygon_convex(int Size = 0){
P.resize(Size);
}
};
bool comp_less(const point &a, const point &b){
return cmp(a.x - b.x) < 0 || cmp(a.x - b.x) == 0 && cmp(a.y - b.y) < 0;
}
polygon_convex convex_hull(vector<point> a){
polygon_convex res(2 * a.size() + 5);
sort(a.begin(), a.end(), comp_less);
a.erase(unique(a.begin(), a.end()), a.end());
int m = 0;
for(int i = 0; i < a.size(); ++i){
while(m > 1 && cmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
--m;
res.P[m++] = a[i];
}
int k = m;
for(int i = int(a.size()) - 2; i >= 0; --i){
while(m > k && cmp(det(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2])) <= 0)
--m;
res.P[m++] = a[i];
}
res.P.resize(m);
if(a.size() > 1) res.P.resize(m - 1);
return res;
}
int T;
double a, b, c;
double w, h;
bool w_h_swap = false;
bool wrong(point r){
return !(cmp(r.x - 0) >= 0 && cmp(w - r.x) >= 0 && cmp(r.y - 0) >= 0 && cmp(h - r.y) >= 0);
}
// #define __DEBUG
int main(){
scanf("%d", &T);
#ifdef __DEBUG
printf("%d\n", T);
#endif
while(T--){
scanf("%lf%lf%lf%lf%lf", &w, &h, &a, &b, &c);
#ifdef __DEBUG
printf("%.0f %.0f %.0f %.0f %.0f ", w, h, a, b, c);
#endif
double l[3] = {a, b, c};
sort(l, l + 3);
w_h_swap = false;
point A, B, C;
double dgr;
A = point(0, 0);
C = point(-1, -1);
if(wrong(C)){
if(l[0] <= w)
B = point(l[0], 0);
else
B = point(w, sqrt(sqr(l[0]) - sqr(w)));
if(!wrong(B)){
dgr = acos( (sqr(l[0]) + sqr(l[1]) - sqr(l[2])) / (2 * l[0] * l[1]) );
C = rotate_point(B, dgr) / B.norm() * l[1];
if(wrong(C)){
dgr = acos( (sqr(l[0]) + sqr(l[2]) - sqr(l[1])) / (2 * l[0] * l[2]) );
C = rotate_point(B, dgr) / B.norm() * l[2];
}
}
}
if(wrong(C)){
if(l[1] <= w)
B = point(l[1], 0);
else
B = point(w, sqrt(sqr(l[1]) - sqr(w)));
if(!wrong(B)){
dgr = acos( (sqr(l[1]) + sqr(l[0]) - sqr(l[2])) / (2 * l[1] * l[0]) );
C = rotate_point(B, dgr) / B.norm() * l[0];
if(wrong(C)){
dgr = acos( (sqr(l[1]) + sqr(l[2]) - sqr(l[0])) / (2 * l[1] * l[2]) );
C = rotate_point(B, dgr) / B.norm() * l[2];
}
}
}
if(wrong(C)){
if(l[2] <= w)
B = point(l[2], 0);
else
B = point(w, sqrt(sqr(l[2]) - sqr(w)));
if(!wrong(B)){
dgr = acos( (sqr(l[2]) + sqr(l[0]) - sqr(l[1])) / (2 * l[2] * l[0]) );
C = rotate_point(B, dgr) / B.norm() * l[0];
if(wrong(C)){
dgr = acos( (sqr(l[2]) + sqr(l[1]) - sqr(l[0])) / (2 * l[2] * l[1]) );
C = rotate_point(B, dgr) / B.norm() * l[1];
}
}
}
if(wrong(C)){
swap(w, h);
w_h_swap = true;
}
if(wrong(C)){
if(l[0] <= w)
B = point(l[0], 0);
else
B = point(w, sqrt(sqr(l[0]) - sqr(w)));
if(!wrong(B)){
dgr = acos( (sqr(l[0]) + sqr(l[1]) - sqr(l[2])) / (2 * l[0] * l[1]) );
C = rotate_point(B, dgr) / B.norm() * l[1];
if(wrong(C)){
dgr = acos( (sqr(l[0]) + sqr(l[2]) - sqr(l[1])) / (2 * l[0] * l[2]) );
C = rotate_point(B, dgr) / B.norm() * l[2];
}
}
}
if(wrong(C)){
if(l[1] <= w)
B = point(l[1], 0);
else
B = point(w, sqrt(sqr(l[1]) - sqr(w)));
if(!wrong(B)){
dgr = acos( (sqr(l[1]) + sqr(l[0]) - sqr(l[2])) / (2 * l[1] * l[0]) );
C = rotate_point(B, dgr) / B.norm() * l[0];
if(wrong(C)){
dgr = acos( (sqr(l[1]) + sqr(l[2]) - sqr(l[0])) / (2 * l[1] * l[2]) );
C = rotate_point(B, dgr) / B.norm() * l[2];
}
}
}
if(wrong(C)){
if(l[2] <= w)
B = point(l[2], 0);
else
B = point(w, sqrt(sqr(l[2]) - sqr(w)));
if(!wrong(B)){
dgr = acos( (sqr(l[2]) + sqr(l[0]) - sqr(l[1])) / (2 * l[2] * l[0]) );
C = rotate_point(B, dgr) / B.norm() * l[0];
if(wrong(C)){
dgr = acos( (sqr(l[2]) + sqr(l[1]) - sqr(l[0])) / (2 * l[2] * l[1]) );
C = rotate_point(B, dgr) / B.norm() * l[1];
}
}
}
if(w_h_swap){
swap(A.x, A.y);
swap(B.x, B.y);
swap(C.x, C.y);
}
point ans[3] = {A, B, C};
point X, Y, Z;
bool flag = false;
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
if(j == i) continue;
for(int k = 0; k < 3; k++){
if(k == i || k == j) continue;
X = ans[i];
Y = ans[j];
Z = ans[k];
if(cmp(dist(X, Y) - a) == 0 &&
cmp(dist(X, Z) - b) == 0 &&
cmp(dist(Y, Z) - c) == 0){
flag = true;
break;
}
}
if(flag) break;
}
if(flag) break;
}
printf("%.8f %.8f %.8f %.8f %.8f %.8f\n", X.x, X.y, Y.x, Y.y, Z.x, Z.y);
}
return 0;
}
来源：oschina
链接：https://my.oschina.net/u/4324366/blog/3442812