标签：
A：
题目链接：点击打开链接
Alice‘s present
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
using namespace std;
int n, m;
int a[501000];
map<int, int> mp;
int main() {
while (scanf("%d", &n) != EOF) {
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
scanf("%d", &m);
int l, r;
for (int j = 1; j <= m; j++){
scanf("%d%d", &l, &r);
int i;
mp.clear();
for ( i = r; i >= l; i--){
mp[a[i]]++;
if (mp[a[i]]>1){ printf("%d\n", a[i]); break; }
}
if (i < l)printf("OK\n");
}
printf("\n");
}
}B：
题目链接：点击打开链接
Bounty hunter
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int N=1e5+9;
double a[N],b[N];
double dp[N],_dp[N];
int main(){
int i,j,k,n,m,T;
double x,y;
while(~scanf("%d%lf%lf",&n,&x,&y)){
for(i=1;i<=n;i++)
scanf("%lf%lf",a+i,b+i);
dp[n+1]=1;
_dp[n+1]=0;
for(i=n;i>=1;i--){
_dp[i]=_dp[i+1]+b[i]*dp[i+1];
dp[i]=max(dp[i+1],_dp[i]/a[i]);
}
printf("%.2lf\n",dp[1]*x+_dp[1]*y);
}
return 0;
}C：
题目链接：点击打开链接
Cinema in Akiba
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX_N = 100007;
int n, m;
int bit[MAX_N], a[MAX_N];
int sum(int i) {
int s = 0;
while (i > 0) {
s += bit[i];
i -= i & -i;
}
return s;
}
void add(int i, int x) {
while (i <= n) {
bit[i] += x;
i += i & -i;
}
}
int main() {
while (1 == scanf("%d", &n)) {
memset(bit, 0, sizeof bit);
for (int i = 1; i <= n; ++i)
add(i, 1);
for (int i = 1; i <= n; ++i) {
int v;
scanf("%d", &v);
int l = 0, r = n;
while (l + 1 < r) {
int mid = (l + r) >> 1;
if (sum(mid) < v) l = mid;
else r = mid;
}
a[i] = r;
add(r, -1);
}
scanf("%d", &m);
for (int i = 0; i < m; ++i) {
int v;
scanf("%d", &v);
if (i == 0) printf("%d", a[v]);
else printf(" %d", a[v]);
}
puts("");
}
return 0;
}
/*
6
4 1 3 3 1 1
6
1 2 3 4 5 6
*/题目链接:点击打开链接
Decode
#include<cstdio>
#include <cstring>
#include<vector>
using namespace std;
int n, k, m;
vector<int> base;
char s[50];
void getbase() {
vector<int> v;
for(int i = n - 1; i >= 0; --i)
for(int j = 0; j < base.size(); ++j)
if(base[j] & (1 << i)) {
int x=base[j];
v.push_back(x);
for(int k=0;k<base.size();k++) if(base[k]&(1<<i)) base[k]^=x;
break;
}
base=v;
}
void out(int x) {
for(int i=n-1;i>=0;i--)
putchar(((1<<i)&x)?49:48);puts("");
}
bool jud(int x) {
int j=0;
for(int i=n-1;i>=0&&x;i--) {
if((1<<i)&x) {
while(j<base.size()&&(!((1<<i)&base[j])||base[j]>=(1<<i+1))) j++;
if(j==base.size()) return 0;
else x^=base[j++];
}
}
return 1;
}
int main() {
while(scanf("%d%d%d",&n,&k,&m)!=EOF) {
base.clear();
for(int i=0; i<k; i++) {
scanf("%s",s);
int x=0;
for(int j=0; j<n; j++) x=(x<<1)|(s[j]-48);
base.push_back(x);
}
getbase();
for(int i=0;i<m;i++){
scanf("%s",s);
int ans=0x7fffffff,diff=100;
int x=0;
for(int j=0;j<n;j++) x=(x<<1)|(s[j]-48);
for(int i=0;i<=n;i++)
for(int j=i;j<=n;j++)
for(int k=j;k<=n;k++) {
int dif=(i!=n)+(j!=n)+(k!=n);
int tmp=(x^(1<<i)^(1<<j)^(1<<k))&((1<<n)-1);
if((dif<diff||(dif==diff&&tmp<ans))&&jud(tmp)) diff=dif,ans=tmp;
}
if(ans==0x7fffffff) puts("NA");
else out(ans);
}
}
return 0;
}E:
题目链接：点击打开链接
Education Manage System
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
const int MAX_N = 200007;
const int MAX_T = 24 * 60 * 400;
const int day[] = {0, 31, 60, 91, 121,
152, 182, 213, 244, 274, 305, 335, 366};
struct Node {
int l, r;
double v;
Node () {
}
Node (int _l, int _r, double _v) {
l = _l, r = _r, v = _v;
}
};
int n;
double dp[MAX_T];
vector<Node> a[MAX_T];
int Month(string s) {
if(s == "Jan") return 1;
if(s == "Feb") return 2;
if(s == "Mar") return 3;
if(s == "Apr") return 4;
if(s == "May") return 5;
if(s == "Jun") return 6;
if(s == "Jul") return 7;
if(s == "Aug") return 8;
if(s == "Sep") return 9;
if(s == "Oct") return 10;
if(s == "Nov") return 11;
if(s == "Dec") return 12;
}
int input() {
char str[10], str2[10];
scanf("%s", str);
int d, h, m;
scanf("%d%*s", &d);
scanf("%d:%d", &h, &m);
scanf("%s", str2);
if(str2[0] == 'p') h += 12;
return m + h * 60 + (d + day[Month(str) - 1]) * 24 * 60;
}
int main() {
while (1 == scanf("%d", &n)) {
memset(dp, 0, sizeof dp);
int ma = 0;
for (int i = 0; i < MAX_T; ++i) a[i].clear();
for (int i = 0; i < n; ++i) {
int l, r;
double v;
l = input();
r = input();
ma = max(r, ma);
scanf("%lf", &v);
// printf("%d %d %f\n", l, r, v);
a[r + 5].push_back(Node(l, r + 5, v));
}
for (int i = 1; i < MAX_T; ++i) {
dp[i] = dp[i - 1];
for (int j = 0; j < a[i].size(); ++j) {
dp[i] = max(dp[i], dp[a[i][j].l] + a[i][j].v);
}
}
printf("%.1f\n", dp[MAX_T - 1]);
}
return 0;
}F：
题目链接：点击打开链接
Fruit Ninja
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <set>
#include <vector>
#include <map>
using namespace std ;
const int INF = 0x3F3F3F3F;
const int MAXN = 1100 ;
const int mod = 100000007 ;
map<string,int> LOW , HIGH ;
int n , m ;
char s1[MAXN] , s2[MAXN] , str[MAXN] ;
char op[MAXN] , s[MAXN] ;
vector<int> l ;
int xx = 0,x=0 ;
long long result = 0 ;
int fact[10000050];
int power(int x,int y){
int res=1;
while (y){
if (y&1) res=(long long)res*x%mod;
x=(long long)x*x%mod;
y>>=1;
}
return res;
}
int inv(int x){
return power(x,mod-2);
}
int C(int x , int y){
if(x>10000020) while(1);
return (long long)fact[x] * inv(fact[x-y]) % mod * inv(fact[y]) % mod ;
}
void dfs(int pos , int now , bool flag){
if (pos == x){
if (now > m) return ;
if (flag) result -= C(n + m - now - 1 , n - 1) ;
else result += C(n + m - now - 1 , n - 1) ;
result %= mod ;
return ;
}
dfs(pos + 1 , now + l[pos] , flag ^ 1) ;
dfs(pos + 1 , now , flag ) ;
}
int main(){
fact[0] = 1;
for (int i = 1 ; i <= 10000030 ; i++) fact[i] = (long long)fact[i-1] * i % mod;
while (~scanf("%d",&n) && n>0){
scanf("%d",&m);
gets(str);
x=0;
xx=0;
bool flag =true;
l.clear() ;
while(1)
{
if (!gets(str)) break;
if (strlen(str) < 2) break;
int X;
sscanf(str, "%s%s%s%d",s1,s1,s2,&X);
if (*s1 == 'l') {l.push_back(X);x++;if(!X) flag=false;}
else {m-=X+1;}
xx++;
}
if (!flag){
printf("0\n");
continue ;
}
if (m < 0) printf("0\n");
else {
result = 0 ;
dfs(0,0,0) ;
printf("%lld\n",(result % mod + mod) % mod) ;
}
}
return 0 ;
}题目链接：点击打开链接
Help Me Escape
算一下最大开门次数，其实增长是一个二次的，所以爆搜就可以了
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
const double P = sqrt(5.0);
int a[N], n;
double dp[20005];
double dfs(int f) {
if(dp[f] > 0) return dp[f];
dp[f] = 0;
for(int i = 0; i < n; i ++) {
if(f > a[i]) {
int num = (int)((P+1)/2.0*a[i]*a[i]);
dp[f] += (double)num/n;
} else {
dp[f] += (1+dfs(f+a[i]))/n;
}
}
return dp[f];
}
int main() {
int f;
while (~scanf("%d%d", &n, &f)) {
memset(dp, 0, sizeof dp);
for(int i = 0; i < n; i ++) {
scanf("%d", &a[i]);
}
printf("%.3f\n", dfs(f));
}
return 0;
}I：
题目链接：点击打开链接
Information Sharing
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if (x>9) pt(x / 10);
putchar(x % 10 + '0');
}
using namespace std;
const int inf = 1e9;
const int N = 100010;
const int M = -1;
int n;
map<string, int>mp;
set<int>st[N], tmp;
set<int>::iterator it;
int f[N], top;
int find(int x){
if (x == f[x])return x;
int y = find(f[x]);
for (it = st[x].begin(); it != st[x].end(); it++)
st[y].insert(*it);
st[x].clear();
return f[x] = y;
}
void Union(int x, int y){
int fx = find(x), fy = find(y);
if (fx == fy)return;
if (fx > fy)swap(fx, fy);
for (it = st[fx].begin(); it != st[fx].end(); it++)
st[fy].insert(*it);
st[fx].clear();
f[fx] = fy;
}
char op[20];
string s, t;
int num, u;
void add(){
cin >> s;
f[top] = top;
st[top].clear();
rd(num);
while (num--){ rd(u); st[top].insert(u); }
mp[s] = top++;
}
void check(){
cin >> s;
int x = find(mp[s]);
pt((int)st[x].size()); putchar('\n');
}
int main() {
while (~scanf("%d", &n)) {
mp.clear();
top = 0;
int out = false;
while (n--){
scanf("%s", op);
if (op[0] == 'a'){
add();
}
else if (op[0] == 's'){
cin >> s; cin >> t;
Union(mp[s], mp[t]);
}
else{
check();
out = true;
}
}
}
return 0;
}/*
99
arr a 3 1 2 3
arr b 1 4
arr c 1 5
arr d 1 6
chec a
sha a b
chec b
chec a
sha a d
chec a
chec d
99
arr a 0
arr b 0
arr c 1 1
c a
c c
sha b c
c b
sha a b
c a
3311 Dig The Wells
4085 Peach Blossom Spring
1970 Ticket to Ride
ZOJ-3613
POJ 3123
*/J：显然每个人的下界是没什么用的，因为可以通过一些和目标点无关的回路来消除，且一定存在这种回路==，所以直接跑最大流即可，目标点所得到的信息由于不受限制，所以直接从源点建边到汇点就好。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if (x>9) pt(x / 10);
putchar(x % 10 + '0');
}
using namespace std;
const int inf = 1e9;
const int N = 5000;
const int M = 500010;
const int INF = ~0u >> 2;
template<class T>
struct Max_Flow {
int n;
int Q[N], sign;
int head[N], level[N], cur[N], pre[N];
int nxt[M], pnt[M], E;
T cap[M];
void Init(int n) {
this->n = n + 1;
E = 0;
std::fill(head, head + this->n, -1);
}
//有向rw 就= 0
void add(int from, int to, T c, T rw) {
pnt[E] = to;
cap[E] = c;
nxt[E] = head[from];
head[from] = E++;
pnt[E] = from;
cap[E] = rw;
nxt[E] = head[to];
head[to] = E++;
}
bool Bfs(int s, int t) {
sign = t;
std::fill(level, level + n, -1);
int *front = Q, *tail = Q;
*tail++ = t; level[t] = 0;
while (front < tail && level[s] == -1) {
int u = *front++;
for (int e = head[u]; e != -1; e = nxt[e]) {
if (cap[e ^ 1] > 0 && level[pnt[e]] < 0) {
level[pnt[e]] = level[u] + 1;
*tail++ = pnt[e];
}
}
}
return level[s] != -1;
}
void Push(int t, T &flow) {
T mi = INF;
int p = pre[t];
for (int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {
mi = std::min(mi, cap[p]);
}
for (int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {
cap[p] -= mi;
if (!cap[p]) {
sign = pnt[p ^ 1];
}
cap[p ^ 1] += mi;
}
flow += mi;
}
void Dfs(int u, int t, T &flow) {
if (u == t) {
Push(t, flow);
return;
}
for (int &e = cur[u]; e != -1; e = nxt[e]) {
if (cap[e] > 0 && level[u] - 1 == level[pnt[e]]) {
pre[pnt[e]] = e;
Dfs(pnt[e], t, flow);
if (level[sign] > level[u]) {
return;
}
sign = t;
}
}
}
T Dinic(int s, int t) {
pre[s] = -1;
T flow = 0;
while (Bfs(s, t)) {
std::copy(head, head + n, cur);
Dfs(s, t, flow);
}
return flow;
}
};
Max_Flow <int>F;
int n, m, l[211], r[210];
set<int>st[205], tmp;
set<int>::iterator it;
vector<int>G;
void input(){
G.clear();
for (int i = 1, num, u; i <= n; i++){
rd(num);
rd(l[i]); rd(r[i]);
st[i].clear();
while (num--){ rd(u); st[i].insert(u); G.push_back(u); }
}
sort(G.begin(), G.end());
G.erase(unique(G.begin(), G.end()), G.end());
rd(m);
for (int i = 1; i <= n; i++){
tmp.clear();
for (it = st[i].begin(); it != st[i].end(); it++)
tmp.insert(lower_bound(G.begin(), G.end(), *it) - G.begin() + 1);
st[i] = tmp;
}
}
int main() {
while (~scanf("%d", &n)) {
input();
int from = 0, to = n + G.size() + 1;
F.Init(to + 10);
for (int i = 1; i <= n; i++){
F.add(from, i, r[i], 0);
for (it = st[i].begin(); it != st[i].end(); it++){
F.add(i, n + (*it), 1, 0);
}
}
F.add(from, m, st[m].size(), 0);
for (int i = 0; i < G.size(); i++)
F.add(n + 1 + i, to, 1, 0);
printf("%d\n", F.Dinic(from, to));
}
return 0;
}题目链接：点击打开链接
Keep Deleting
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char a[257], b[512005];
int main() {
while (~scanf("%s", a)) {
int l = strlen(a);
scanf("%s", b);
int len = strlen(b), size = l-1, cnt = 0;
for(int i = l-1; i < len; i ++) {
b[size++] = b[i];
if(size >= l) {
int ok = 1;
for(int j = 0; j < l; j ++) {
if(a[j] != b[size-l+j]) {
ok = 0;
break;
}
}
if(ok) {
cnt ++;
size -= l;
}
}
}
printf("%d\n", cnt);
}
return 0;
}标签：
原文地址：http://blog.csdn.net/qq574857122/article/details/43968135
