UCF Local Programming Contest 2012（Practice）
A. Wall Street Monopoly
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思路：(cost = cost[i][j]+cost[j+1][j+i]+100*min(len[j][j],dep[j][j])*min(len[j+1][j+i],dep[j+1][j+i]))
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 25;
const int times = 100;
const int INF = 0x3f3f3f3f;
struct Dim{
int len, dep, cost;
Dim(){ }
Dim(int len_, int dep_, int cost_): len(len_), dep(dep_), cost(cost_){ }
};
int t, n, len, dep, cost;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
for (int tot = 1; tot <= t; tot ++ ){
cin >> n;
Dim dim[N];
for (int i = 0; i < n; i ++ ){
cin >> dim[i].len >> dim[i].dep;
dim[i].cost = 0;
}
Dim ans[N][N];
for (int i = 0; i < n; i ++ )
ans[i][i] = dim[i];
for (int i = 1; i < n; i ++ ){
for (int j = 0; j < n - i; j ++ ){
len = ans[j][j].len + ans[j + 1][j + i].len;
dep = max(ans[j][j].dep, ans[j + 1][j + i].dep);
cost = ans[j][j].cost + ans[j + 1][j + i].cost + times * min(ans[j][j].len, ans[j][j].dep) * min(ans[j + 1][j + i].len, ans[j + 1][j + i].dep);
// cout << len << " " << dep << " " << cost << "
";
Dim min_(len, dep, cost);
for (int k = j + 1; k < j + i; k ++ ){
len = ans[j][k].len + ans[k + 1][j + i].len;
dep = max(ans[j][k].dep, ans[k + 1][j + i].dep);
cost = ans[j][k].cost + ans[k + 1][j + i].cost + times * min(ans[j][k].len, ans[j][k].dep) * min(ans[k + 1][j + i].len, ans[k + 1][j + i].dep);
// cout << len << " " << dep << " " << cost << "
";
if (cost < min_.cost){
min_.len = len;
min_.dep = dep;
min_.cost = cost;
}
}
ans[j][j + i] = min_;
}
}
// for (int i = 0; i < n; i ++ ){
// for (int j = 0; j < n; j ++ )
// cout << ans[i][j].cost << " ";
// cout << "
";
// }
cout << "The minimum cost for lot #" << tot << " is $" << ans[0][n - 1].cost << ".
";
}
return 0;
}
B. How Old Are You Mr.String?
-
思路：字符串处理水题
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int t;
string a, b;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
for (int i = 1; i <= t; i ++ ){
int cnt1[26] = {0}, cnt2[26] = {0};
cout << "Data set #" << i << ": ";
cin >> a >> b;
for (auto c: a)
cnt1[c - 'a'] ++ ;
for (auto c: b)
cnt2[c - 'a'] ++ ;
bool flag1 = false, flag2 = false;
for (int j = 25; j >= 0; j -- ){
if (cnt1[j] > cnt2[j]){
flag1 = true;
break;
}
else if (cnt1[j] < cnt2[j]){
flag2 = true;
break;
}
else
continue;
}
if (flag1)
cout << "First string is older
";
else if (flag2)
cout << "First string is younger
";
else
cout << "The two strings are the same age
";
}
return 0;
}
C. Clean Up the Powers that Be
-
思路：模拟题
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 10010;
int t, n, base, exp_;
int prime[N];
int get_len(int x){
int len = 0;
while (x){
x /= 10;
len ++ ;
}
return len;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
for (int tot = 1; tot <= t; tot ++ ){
cin >> n;
memset(prime, 0, sizeof(prime));
for (int i = 1; i <= n; i ++ ){
cin >> base >> exp_;
prime[base] += exp_;
}
cout << "Prime Factorization #" << tot << ":
";
for (int i = 2; i < N; i ++ ){
if (!prime[i])
continue;
else{
int len = get_len(i);
for (int i = 1; i <= len; i ++ )
cout << " ";
cout << prime[i];
}
}
cout << "
";
for (int i = 2; i < N; i ++ ){
if (!prime[i])
continue;
else{
cout << i;
int len = get_len(prime[i]);
for (int i = 1; i <= len; i ++ )
cout << " ";
}
}
cout << "
";
}
return 0;
}
D. The Clock Algorithm
-
思路：阅读理解题 同样的读完题模拟
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 60;
int n, r, p, r_, k, now, c;
int v[N];
bool old[N];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
while (cin >> n >> r && n){
cout << "Program " << ++ p << "
";
memset(v, -1, sizeof(v));
memset(old, false, sizeof(old));
now = 0, k = 0;
while (r -- ){
cin >> r_;
c = -1;
for (int i = 0; i < n; i ++ ){
if (v[i] == r_){
c = i;
break;
}
}
if (c >= 0){
cout << "Access page " << r_ << " in cell " << c + 1 << ".
";
old[c] = false;
continue;
}
k ++ ;
c = -1;
for (int i = 0; i < n; i ++ ){
if (v[i] < 0){
c = i;
break;
}
}
if (c >= 0){
cout << "Page " << r_ << " loaded into cell " << c + 1 << ".
";
v[c] = r_;
old[c] = false;
continue;
}
while (true){
if (old[now]){
cout << "Page " << r_ << " loaded into cell " << now + 1 << ".
";
v[now] = r_;
old[now] = false;
now = (now + 1) % n;
break;
}
else{
old[now] = true;
now = (now + 1) % n;
}
}
}
cout << "There are a total of " << k << " page faults.
";
}
return 0;
}
G. Lifeform Detector
-
思路：S后面可以跟空格 每个a必须有b相对应 c既可以当S也可以当T
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int t;
string s;
bool judge(string s){
stack<char> st;
for (auto a: s){
if (a == 'a')
st.push('a');
else if (a == 'b'){
if (!st.empty())
st.pop();
else
return false;
}
else if (a != 'c')
return false;
}
if (!st.empty())
return false;
else
return true;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
for (int tot = 1; tot <= t; tot ++ ){
cin >> s;
cout << "Pattern " << tot << ": ";
if (judge(s))
cout << "More aliens!
";
else
cout << "Still Looking.
";
}
return 0;
}
H. Ordered Numbers
-
思路：水题
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int t;
int a[3];
int main(){
#ifndef ONLINE_JUDGE
freopen("order.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
for (int i = 1; i <= t; i ++ ){
cin >> a[0] >> a[1] >> a[2];
cout << "Data set #" << i << ":
";
cout << " Original order: " << a[0] << " " << a[1] << " " << a[2] << "
";
sort(a, a + 3);
cout << " Smallest to largest: " << a[0] << " " << a[1] << " " << a[2] << "
";
}
return 0;
}