2013 Phuket Regional
A Spanning trees in a secure lock pattern
用矩阵运算计算生成树数量，题目已经给出计算方式，只要按着算一遍...
有兴趣的可以去看看Matrix-Tree定理(Kirchhoff矩阵-树定理)
#include
#include
#include
#include
#include
using namespace std;
const int N = 41;
int fx[8][2]={
{1,1},{1,-1},{-1,1},{-1,-1},
{1,0},{0,-1},{-1,0},{0,1}};
long long Mat[N][N];
int n;
long long Gauss(int n){
long long ans=1,t,sav;
int i,j,k;
for(int i=1;i=0 && tx=0 && ty B Teaching Hazard
#include
#include
#include
#define maxn 100100
typedef long long ll;
using namespace std;
ll a[maxn],pri[maxn],prime=0;
ll top,ans,n,x,tmp;
bool b[maxn]={false};
void getpri()
{
for (int i=2; i0) a[i]+=(tmp=tmp/pri[i]);
if (a[i]==0)
{
top=i-1;
return;
}
}
}
ll getnum(ll X)
{
ll tot=1,cur;
for (int i=1; i<=top && a[i]>=X; i++) tot*=a[i]/X+1;
return tot;
}
int main()
{
getpri();
while (scanf("%lld%lld",&n,&x) && (n|x) )
{
init();
ll pre,next=getnum(x);
while (true)
{
x++;
pre=getnum(x);
ans+=(next-pre)*(next-pre-1)/2;
if (pre==1) break;
swap(pre,next);
}
printf("%lld\n",ans);
}
return 0;
} C Counting Lattice Squares
#include
#include
#include
#define MAXN 1000010
typedef long long LL;
LL A[MAXN], B[MAXN], C[MAXN], a[MAXN];
int main() {
memset(a, 0, sizeof(a));
for(int i = 1; i <= 1000000; i += 2) {
a[i] = i;
}
A[0] = B[0] = C[0] = 0;
for(int i = 1; i <= 1000000; i ++) {
A[i] = A[i - 1] + a[i];
B[i] = B[i - 1] + a[i] * i;
C[i] = C[i - 1] + a[i] * i * i;
}
int n, m;
while(scanf("%d%d", &n, &m), n) {
if(n > m) std::swap(n, m);
LL ans = A[n] * (n + 1) * (m + 1) - B[n] * (n + 1 + m + 1) + C[n];
printf("%lld\n", ans);
}
return 0;
} D Seven Segment Display
该题题意学过数电的同学应该很快就能看懂...不过和数电没什么关系= =
就是给你一个图要你判断它是否能组成给出的结构，代码题
首先将0~9分类，0第一类，1、2、5、7第二类，8第三类，6、9第四类，3、4第五类
然后就是计算图的度，根据要求把图分割求链的长度，然后判断是否相等即可
#include
#include
#include
#include
#include
using namespace std;
const int N = 510, M = 1010;
vector< pair > ans;
struct Edge{
int next,to;
} e[M*2];
int n,m,tot;
int head[N],de[N],dep[4],ret,res;
bool vis[N];
void add_edge(int a,int b){
e[tot].to=b;e[tot].next=head[a];
head[a]=tot++;
}
void dfs(int p){
vis[p]=true;
for(int i=head[p];~i;i=e[i].next)
if(!vis[e[i].to]) dfs(e[i].to);
}
int dfs2(int p,int fa){
if(p==ret) return 0;
int res;
for(int i=head[p];~i;i=e[i].next)
if(e[i].to!=fa) res=dfs2(e[i].to,p)+1;
return res;
}
int main()
{
int T,cas=1,a,b;scanf("%d",&T);
for(cas=1;cas<=T;++cas){
printf("Case %d: ",cas);
scanf("%d%d",&n,&m);tot=0;
ans.clear();
memset(head,-1,sizeof head);
memset(de,0,sizeof de);
memset(dep,0,sizeof dep);
memset(vis,0,sizeof vis);
for(int i=0;i3 || de[i]==0){
ok=false;
break;
}
++dep[de[i]];
}
if(dep[3]>2) ok=false;
if(!ok){
printf("0\n");
if(cas!=T) printf("\n");
continue;
}
if(dep[2]==n && n>=6){
if((n-6) % 6 == 0) ans.push_back(make_pair(0,n/6-1));
}
else if(dep[1]==2 && dep[2]==n-2){
if((n-3) % 2 == 0) ans.push_back(make_pair(1,(n-3)/2));
if((n-6) % 5 == 0) ans.push_back(make_pair(2,(n-6)/5)),ans.push_back(make_pair(5,(n-6)/5));
if((n-4) % 3 == 0) ans.push_back(make_pair(7,(n-4)/3));
}
else if(dep[3]==2 && dep[2]==n-2){
if((n-6) % 7 == 0) ans.push_back(make_pair(8,(n-6)/7));
}
else if(dep[3]==1 && dep[1]==1 && dep[2]==n-2){
for(int i=1;i<=n;++i){
if(de[i]==1) res=i;
if(de[i]==3) ret=i;
}
int l=dfs2(res,-1);
int last=n-l-1;
if(l % 2 == 0) l=(l-2)/2;
if((last-3) % 4 == 0) last=(last-3)/4;
if(l==last) ans.push_back(make_pair(6,l)),ans.push_back(make_pair(9,l));
}
else if(dep[3]==1 && dep[1]==3 && dep[2]==n-4){
vector ii,ll;
for(int i=1;i<=n;++i){
if(de[i]==3) ret=i;
if(de[i]==1) ii.push_back(i);
}
for(int i=0;i<3;++i)
ll.push_back(dfs2(ii[i],-1));
sort(ll.begin(),ll.end());
if(ll[2]==2*ll[0]){
if(ll[0]==ll[1]) ans.push_back(make_pair(4,ll[0]-1));
if(ll[1]==ll[2]) ans.push_back(make_pair(3,ll[0]-1));
}
}
else ok=false;
sort(ans.begin(),ans.end());
printf("%d\n",ans.size());
for(int i=0;i E Airport Sort
#include
#include
#include
#define MAXN 20010
struct P {
int tic, no, loc;
bool operator < (const P &t) const {
if(no != t.no) return no < t.no;
return loc < t.loc;
}
}p[MAXN];
int N, K;
void input() {
int x;
scanf("%d%d", &N, &K);
for(int i = 1; i <= N; i ++) {
scanf("%d", &x);
p[i].tic = x, p[i].no = (x - 1) / K, p[i].loc = i;
}
std::sort(p + 1, p + 1 + N);
}
int sum[4 * MAXN], D;
void build() {
for(D = 1; D < N + 2; D <<= 1);
for(int i = 0; i < D; i ++) {
if(i >= 1 && i <= N) sum[D + i] = 1;
else sum[D + i] = 0;
}
for(int i = D - 1; i >= 1; i --) {
sum[i] = sum[i << 1] + sum[i << 1 | 1];
}
}
void update(int i) {
for(i >>= 1; i > 0; i >>= 1) {
sum[i] = sum[i << 1] + sum[i << 1 | 1];
}
}
int query(int x, int y) {
int ans = 0;
for(int i = D + x - 1, j = D + y + 1; i + 1 != j; i >>= 1, j >>= 1) {
if(~i & 1) ans += sum[i ^ 1];
if(j & 1) ans += sum[j ^ 1];
}
return ans;
}
void process() {
int X = 0, Y = 0;
for(int i = 1; i <= N; i ++) {
Y = std::max(Y, std::abs(p[i].loc - i));
}
build();
for(int i = 1; i <= N; i ++) {
if(p[i].loc > 1) X += query(1, p[i].loc - 1);
sum[D + p[i].loc] = 0, update(D + p[i].loc);
}
printf("%d\n", X - Y);
}
int main() {
//freopen("test.in", "rb", stdin);
int t;
scanf("%d", &t);
for(int tt = 1; tt <= t; tt ++) {
input();
printf("Case %d: ", tt);
process();
}
return 0;
} F Boxes
#include
#include
#include
#define MOD 10007
typedef long long LL;
struct Box {
int w, n;
bool operator < (const Box &t) const {
return w > t.w;
}
}b[110];
int B;
int N, W;
int f[110][110][1010];
int fac[210], ni[210], facNi[210];
void input() {
scanf("%d%d", &N, &W);
B = 0;
for(int i = 0; i < N; i ++) {
int w;
scanf("%d", &w);
int id = -1;
for(int j = 1; j <= B; j ++) {
if(b[j].w == w) {
id = j;
break;
}
}
if(id == -1) {
++ B;
b[B].w = w;
b[B].n = 1;
} else {
++ b[id].n;
}
}
}
void prepare() {
std::sort(b + 1, b + 1 + B);
for(int i = 0; i <= B; i ++) {
for(int j = 0; j <= N; j ++) {
for(int k = 0; k <= W; k ++) {
f[i][j][k] = -1;
}
}
}
f[0][0][0] = 1;
for(int i = 1; i <= B; i ++) {
for(int j = 0; j <= N; j ++) {
for(int k = 0; k <= W; k ++) {
bool ok = false;
int ans = 0;
if(f[i - 1][j][k] != -1) {
ok = true;
ans += f[i - 1][j][k];
}
for(int m = 1; m <= b[i].n; m ++) {
int pj = j - m, pk = k - b[i].w * m;
if(pj < 0 || pk < 0) break;
if(f[i - 1][pj][pk] != -1) {
ok = true;
ans += ((LL)fac[pj + m] * facNi[m] * facNi[pj] * f[i - 1][pj][pk]) % MOD;
if(ans >= MOD) ans -= MOD;
}
}
if(ok) f[i][j][k] = ans;
else f[i][j][k] = -1;
}
}
}
}
int power(int A,int B)
{
int tot=1;
while (B)
{
if (B&1) tot=(tot*A)%MOD;
A=(A*A)%MOD;
B>>=1;
}
return tot;
}
void process() {
int ans = 0;
for(int r = 0; r < W; r ++) {
int t = 1, n = 0, space = 0;
int i = B;
for(; i >= 1; i --) {
if(b[i].w > r) break;
t = ((LL)fac[n + b[i].n] * facNi[n] * facNi[b[i].n] * t) % MOD;
space += b[i].n * b[i].w;
n += b[i].n;
}
if(space <= W - r) {
space = W - r - space;
int sum = 0;
for(int j = 0; j <= N; j ++) {
if(f[i][j][space] != -1) {
sum += ((LL)fac[n + j] * facNi[n] * facNi[j] * f[i][j][space]) % MOD;
if(sum >= MOD) sum -= MOD;
}
}
ans += (t * sum) % MOD;
if(ans >= MOD) ans -= MOD;
}
}
printf("%d\n", ans);
}
int main() {
fac[0] = facNi[0] = 1;
for(int i = 1; i <= 200; i ++) {
int ni = power(i, MOD - 2);
fac[i] = fac[i - 1] * i % MOD;
facNi[i] = facNi[i - 1] * ni % MOD;
}
int t, tt;
scanf("%d", &t);
for(tt = 1; tt <= t; tt ++) {
input();
prepare();
printf("Case %d: ", tt);
process();
}
return 0;
}
G Meeting Room Arrangement
#include
#include
#include
#include
using namespace std;
struct Node{
int s,f;
} q[30];
int n,tot,ans;
bool cmp(Node a,Node b){
return a.s==b.s ? a.f=q[o].f) dfs(i,res+1);
}
int main()
{
//freopen("test.in","rb",stdin);
scanf("%d",&n);
while(n--){
tot=0;
while(scanf("%d%d",&q[tot].s,&q[tot].f),q[tot].s+q[tot].f) ++tot;
sort(q,q+tot,cmp);++tot;
ans=0;
for(int i=0;i H
I Cabin Baggage
#include
#include
#include
#include
using namespace std;
int n,ans;
int main()
{
//freopen("test.in","rb",stdin);
scanf("%d",&n);
while(n--){
double l,w,d,we;
scanf("%lf%lf%lf%lf",&l,&w,&d,&we);
bool ok=true;
if(l<=56.00 && w<=45.00 && d<=25.00) ok=true;
else if(l+w+d <= 125.00) ok=true;
else ok=false;
if(we>7) ok=false;
if(ok) puts("1"),++ans;
else puts("0");
}
printf("%d\n",ans);
return 0;
} J Minimal Subarray Length
#include
#include
#include
#define MAXN 500010
typedef long long LL;
int N, a[MAXN], X;
int D, tree[4 * MAXN];
LL A[MAXN];
int M;
void build() {
A[0] = 0;
for(int i = 1; i <= N; i ++) {
A[i] = a[i] + A[i - 1];
}
std::sort(A + 1, A + 1 + N);
M = std::unique(A + 1, A + 1 + N) - (A + 1);
for(D = 1; D < M + 2; D <<= 1);
memset(tree, 0, sizeof(tree[0]) * 2 * D);
}
int findId(LL sum) {
int min = 0, max = M + 1, mid;
for(;;) {
mid = (min + max) >> 1;
if(mid == min) break;
if(A[mid] <= sum) min = mid;
else max = mid;
}
return mid;
}
void update(int i) {
for(i >>= 1; i > 0; i >>= 1) {
tree[i] = std::max(tree[i << 1], tree[i << 1 | 1]);
}
}
int query(int x, int y) {
int ans = 0;
for(int i = D + x - 1, j = D + y + 1; i + 1 != j; i >>= 1, j >>= 1) {
if(~i & 1) ans = std::max(ans, tree[i ^ 1]);
if(j & 1) ans = std::max(ans, tree[j ^ 1]);
}
return ans;
}
void process() {
int ans = N + 1;
LL sum = 0;
for(int i = 1; i <= N; i ++) {
sum += a[i];
if(a[i] >= X) ans = 1;
if(sum >= X) ans = std::min(ans, i);
LL x = sum - X;
int id = findId(x);
if(id > 0) {
int j = query(1, id);
if(j > 0 && i - j < ans) ans = i - j;
}
id = findId(sum);
tree[D + id] = i, update(D + id);
}
printf("%d\n", ans <= N ? ans : -1);
}
int main() {
//freopen("test.in", "rb", stdin);
int t;
scanf("%d", &t);
while(t --) {
scanf("%d%d", &N, &X);
for(int i = 1; i <= N; i ++) {
scanf("%d", &a[i]);
}
build();
process();
}
return 0;
}