B:
Uva: 12582 - Wedding of Sultan
给定一个字符串（仅由大写字母构成）一个字母表示一个地点，经过这个点或离开这个点都输出这个地点的字母）
问：
每个地点经过的次数（维护一个栈就可以了，注意进入起点和离开起点都不算入起点的次数）
#include<cstdio>
#include<cstring>
#include<stack>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 105;
int b[N];
char a[N];
stack<char> s;
void work() {
memset(b, 0, sizeof b);
while(s.size()) s.pop();
int n = strlen(a);
s.push(a[0]);
for(int i = 1; i < n; i ++) {
b[s.top() - 'A'] ++;
if(s.top() == a[i]) {
s.pop();
} else {
s.push(a[i]);
}
}
b[a[0]-'A'] --;
for(int i = 0; i < 26; i ++) {
if(b[i] > 0) {
printf("%c = %d\n", i+'A', b[i]);
}
}
}
int main() {
int T, cas = 0;
scanf("%d", &T);
while(T-- > 0) {
scanf("%s", a);
printf("Case %d\n", ++cas);
work();
}
return 0;
}
C:
Uva: 12583 - Memory Overflow
有n天，常数k，n长的字符串
每天有一个字母来拜访主角，主角只能记住最后k天的字母，问主角能辨认出几个字母
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <bits/stdc++.h>
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 N = 1000;
typedef long long ll;
int n, k;
int cnt[N];
char s[N];
int work(){
if(k == 0) return 0;
queue<char> q;
memset(cnt, 0, sizeof cnt);
int ans = 0;
for(int i = 0; s[i]; i++){
if(cnt[s[i]])
ans++;
if((int)q.size() == k)
{
cnt[q.front()]--;
q.pop();
}
q.push(s[i]);
cnt[s[i]]++;
}
return ans;
}
int main(){
int T, Cas = 1; rd(T);
while (T--){
rd(n); rd(k); scanf("%s", s);
int ans = work();
printf("Case %d: %d\n", Cas++, ans);
}
return 0;
}
E:
Uva: 12585 - Poker End Games
有2个人X,Y，每个人手里有纸牌对应a 和 b张
每局游戏有0.5概率 X 获得Y手中 min(a,b)张
0.5概率 Y获得X手中 min(a,b)张
若有人手里没有牌则另一个人胜利。
问获胜需要的场数的期望 和 X获胜的概率
思路：
爆搜一下
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
double E, P;
void dfs(int n, int m, int t) {
if(t > 50) return ;
if(n == 0 || m == 0) {
double p = 1;
for(int i = 0; i < t; i ++) {
p *= 0.5;
}
E += t*p;
if(n != 0) P += p;
} else {
dfs(n-min(n, m), m+min(n, m), t+1);
dfs(n+min(n, m), m-min(n, m), t+1);
}
}
int main() {
int T, cas = 0;
scanf("%d", &T);
while(T-- > 0) {
int n, m;
scanf("%d%d", &n, &m);
E = 0.0, P = 0.0;
dfs(n, m, 0);
printf("Case %d: %.6f %.6f\n", ++cas, E, P);
}
return 0;
}
F:
UVA: 12586 - Overlapping Characters
大模拟，写写写
#include<cstdio>
#include<cstring>
#include<stack>
#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
const int N = 50;
const int r = 17;
const int c = 43;
struct node {
char s[r][c];
};
map<char, node> dic;
int n, q, len;
int vis[r][c];
char ok[N], s[N];
void pu() {
for (int i = 0; i < len; ++i)
ok[i] = 'N';
memset(vis, -1, sizeof vis);
for (int i = 0; i < len; ++i)
for (int j = 0; j < r; ++j)
for (int k = 0; k < c; ++k)
if (dic[s[i]].s[j][k] == '*') {
if (vis[j][k] == -1)
vis[j][k] = i;
else
vis[j][k] = -2;
}
for (int i = 0; i < r; ++i)
for (int j = 0; j < c; ++j)
if (vis[i][j] >= 0)
ok[vis[i][j]] = 'Y';
}
void work() {
node in;
dic.clear();
scanf("%s", s);
len = strlen(s);
for (int i = 0; i < len; ++i) {
for (int j = 0; j < 17; ++j)
scanf("%s", in.s[j]);
dic[s[i]] = in;
}
for (int i = 1; i <= q; ++i) {
scanf("%s", s);
len = strlen(s);
printf("Query %d: ", i);
pu();
for (int j = 0; j < len; ++j)
putchar(ok[j]);
putchar('\n');
}
}
int main() {
while (~scanf("%d%d", &n, &q))
work();
return 0;
}
G:
Uva: 12587 - Reduce the Maintenance Cost
==写死爹了
题解：点击打开链接
I:
Uva: 12589 - Learning Vector
题意：给定n个向量，选k个向量，使得向量围成的面积最大。思路：设我们在 向量a, b 之中选一个。则得到一个方程表示2个向量各自增加的面积化简后就能排个序。
#include<cstdio>
#include<cstring>
#include<stack>
#include<iostream>
#include<algorithm>
using namespace std;
const int Inf = 1e9;
const int H = 2500+2;
const int N = 50+2;
struct node {
int w, h;
};
int T = 0, d[N][N][H];
node a[N];
bool cc(const node& i, const node& j) {
return i.h*j.w>= j.h*i.w;
}
void work() {
int n, K, mxh = 0, s = 0;
scanf("%d%d", &n, &K);
for (int i=0; i<n; ++i) {
scanf("%d%d", &a[i].w, &a[i].h);
s += a[i].h;
}
sort(a, a + n, cc);
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= K; ++j)
for (int k = 0; k <= s; ++k)
d[i][j][k] = -Inf;
d[0][0][0] = 0;
for (int i = 0; i < n; ++i)
for (int j = 0; j <= i && j <= K; ++j)
for (int k = 0; k <= mxh; ++k)
if (d[i][j][k] >= 0) {
d[i+1][j][k] = max(d[i+1][j][k], d[i][j][k]);
if (j+1 <= K) {
d[i+1][j+1][k+a[i].h] = max(d[i+1][j+1][k+a[i].h], d[i][j][k] + a[i].w*k*2 + a[i].h*a[i].w);
mxh = max(mxh, k+a[i].h);
}
}
int ans = 0;
for (int i = 0; i <= mxh; ++i)
ans = max(ans, d[n][K][i]);
printf("Case %d: %d\n", ++T, ans);
}
int main() {
int cas;
scanf("%d", &cas);
while (cas-->0)
work();
return 0;
}
时间： 2024-10-15 20:02:15