2013 Asia Chengdu Regional Contest
B. Fibonacci Tree
- 解法：
- 先跑一遍最小生成树，答案就是需要白边的最小的数量；然后再跑一遍最大生成树，答案就是需要白边的最大的数量。那么看看两个之间有没有夹着一个斐波那契数（这个可以用二分查找找出）。
- 最后要判断图是否连通。这个可以看看最小生成树是否连了 N-1 条边。如果不是的话，图不连通。
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<functional>
using namespace std;
const int maxn = 100010;
int N, M, kase, p[maxn], fib[50] = {
1, 1};
struct P{
int u, v, w;
bool operator < (const P& rhp)const{
return w < rhp.w;
}
bool operator > (const P& rhp)const {
return w > rhp.w;
}
}G[maxn];
void init(int N){
for(int i = 1; i <= N; i++) p[i] = i;
}
int find(int x){
if(p[x] == x) return x;
return p[x] = find(p[x]);
}
void unite(int a, int b) {
if (find(a) == find(b)) return;
p[find(a)] = find(b);
}
int mst(){
init(N);
int res = 0, cnt = 0;
for(int i = 0; i < M; i++){
int u = G[i].u, v = G[i].v, w = G[i].w;
if(find(u) == find(v)) continue;
unite(u, v);
res += w;
cnt++;
}
if(cnt != N - 1) return -1;
return res;
}
int main(){
for(int i = 2; i < 30; i++){
fib[i] = fib[i - 1] + fib[i - 2];
}
int T;
scanf("%d", &T);
while(T--){
scanf("%d%d", &N, &M);
for(int i = 0; i < M; i++){
scanf("%d%d%d", &G[i].u, &G[i].v, &G[i].w);
}
sort(G, G + M);
int minimum = mst();
sort(G, G + M, greater<P>());
int maximum = mst();
bool flag = true;
if(minimum == -1 || maximum == -1){
flag = false;
}
else{
int id1 = lower_bound(fib, fib + 30, minimum) - fib, id2 = lower_bound(fib, fib + 30, maximum) - fib;
if(id1 == id2) flag = false;
}
if(flag) printf("Case #%d: Yes\n", ++kase);
else printf("Case #%d: No\n", ++kase);
}
return 0;
}
J. Just Random
这个讲的是我见到的比较好的题解：传送门
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a, b, c, d, p, m, kase;
ll gcd(ll a, ll b){
if(b == 0) return a;
return gcd(b, a % b);
}
ll cal(ll x, ll y){
if (x < 0 || y < 0) return 0;
x++, y++;
ll A = x / p, B = y / p;
ll C = x % p, D = y % p;
//只要选一个数，就必然可以在另一个集合中选另一个数，使得两者之和模p为m
ll res = p * A * B + A * D + B * C;
C--, D--;
if(C > D) swap(C, D);
for(ll i = m; i <= C + D; i += p){
if(i < C) res += i + 1;
else if(C <= i && i <= D) res += C + 1;
else res += C + D - i + 1;
}
return res;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
scanf("%lld%lld%lld%lld%lld%lld", &a, &b, &c, &d, &p, &m);
ll numerator = cal(b, d) - cal(a - 1, d) - cal(b, c - 1) + cal(a - 1, c - 1);
ll denominator = (b - a + 1) * (d - c + 1);
ll divisor = gcd(numerator, denominator);
printf("Case #%lld: %lld/%lld\n", ++kase, numerator / divisor, denominator / divisor);
}
return 0;
}