PTA甲级—常用技巧与算法
分类:
IT文章
•
2022-04-10 21:24:25

二分
1010 Radix (25分)
题目的大意是,给你两个数,N1和N2,给定其中一个数的基数,然后求另一个数的基数,使得两个数相等
这题有的地方没有给清楚,感觉有点小问题。给定标签的数不会溢出,而未给定标签的数是会溢出long long的,这点题目没有直接说明,溢出的部分需要用<0去判断。再就是二分的下限取小了,反而WA。
这里主要是对二分法的运用,我自己有个点是没有注意到的就是,求这个数的基数,如何取其上下限。如果这个数大于一位,当这个数最小为10时可以取到最大的基数即这个数的十进制,最小就是其各位的最大数字+1.但是当这个数字只有1位的时候,套用上述的方法显然就会出错,这样取得上限比下限还小,这种情况稍微处理一下就好了。
这会借着还复习了下二分法和快速幂(倍增思想(由于事先知道跳跃步数,采用二进制分解,边计算边跳跃的方式)和分治思想(将问题划分为规模更小的子问题))
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
#include <cmath>
#define ll long long
#define inf 0x3f3f3f
using namespace std;
string s1, s2;
ll id, radix;
ll qpow(ll a, ll n){
ll ans = 1;
while(n){
if(n&1) ans *= a; //ans乘上当前的a
a *= a; //a自乘
n >>= 1; //n往右移一位
}
return ans;
}
ll handle(string s, ll d){
ll len = s.length(), num;
ll sum = 0;
for(int i = 0; i < len; i++){
if(s[i]<='9') num = s[i]-'0';
else num = s[i]-'a'+10;
sum += num*qpow(d, len-1-i);
}
return sum;
}
int main(){
cin >> s1 >> s2 >> id >> radix;
if(id==2) swap(s1, s2);
//cout << s1 << " " << s2;
ll tmp = handle(s1, radix);
//二分+快速幂
//char c=*max_element(str.begin(),str.end());
int cnt = -1;
for(int i = 0; s2[i]; i++){
if(s2[i]<='9') cnt = max(cnt, s2[i]-'0');
else cnt = max(cnt, s2[i]-'a'+10);
}
ll l = cnt, r = max(l, tmp)+1;
while(r-l>1){
ll mid = (l+r)/2;
ll res = handle(s2, mid);
if(res>=tmp||res<0) r = mid;
else l = mid;
}
if(handle(s2, r)==tmp) printf("%lld
", r);
else printf("Impossible
");
}
View Code
reference:
https://segmentfault.com/a/1190000037525090
https://blog.****.net/CV_Jason/article/details/80993283
https://blog.****.net/d891320478/article/details/8303072
https://blog.****.net/weixin_30782331/article/details/98610783
1048 Find Coins (25 分)
排序后对于某个a[i]在数列中使用lower_bound查找m-a[i],不过需要注意:查找的位置应该在i以后,满足相加等于m即是最小的V1
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
int n, m, a[maxn];
int main(){
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
sort(a+1, a+1+n);
for(int i = 1; i <= n; i++){
if(a[i]>m/2) break;
int x = a[i], y = *lower_bound(a+1, a+1+n, m-a[i]);
if(x+y==m) return !printf("%d %d", x, y);
}
printf("No Solution");
}
View Code
1044 Shopping in Mars (25 分)
lower_bound的应用
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define pii pair<int, int>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1e5+100;
int n, m, num, sum[maxn], t, lost;
pii sta[maxn];
int main(){
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++)
scanf("%d", &num), sum[i] = sum[i-1] + num;
sum[n+1] = inf, lost = 1e8;
for(int i = 0; i <= n-1; i++){
int head = i+1, tail = lower_bound(sum+1, sum+1+n, sum[i]+m)-sum;
if(sum[tail]-sum[head-1]<lost) lost = sum[tail]-sum[head-1], t = 0, sta[++t] = make_pair(head, tail);
else if(sum[tail]-sum[head-1]==lost) sta[++t] = make_pair(head, tail);
}
sort(sta+1, sta+1+t);
for(int i = 1; i <= t; i++) printf("%d-%d
", sta[i].first, sta[i].second);
}
View Code
1085 Perfect Sequence (25 分)
upper_bound的应用,注意细节和数据范围,应当使用long long
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
ll n, p, a[maxn], res;
int main(){
scanf("%lld%lld", &n, &p);
for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
sort(a+1, a+1+n);
for(int i = 1; i <= n; i++) res = max(res, 1ll*(upper_bound(a+i+1, a+1+n, a[i]*p)-a-i));
printf("%lld", res);
}
View Code
散列
1078 Hashing (25 分)
Quadratic probing (with positive increments only) is used to solve the collisions.
这句话说的是使用平方探测法来解决哈希冲突,Linear Probing(线性探测法)、Quadratic probing(平方探测法)这种专业术语在平常的学习中应当认真记忆而不是认为不重要,因为这句话一开始看不懂,想当然认不重要就略过了,那结果多半WA。
知道了解决办法之后,需要处理的一个问题就是我们如何知道插入失败?
假设$x<y$,由$h(k) + x^2 = h(k) + y^2 quad (mod p)$得:
$$ x^2 = y^2 quad(mod p)$$
$$ (x-y)(x+y) = 0 quad(mod p)$$
由上述式子推导可发现$p$是一个循环节,如果从$0 sim p-1$进行枚举仍然找不到位置的话即可认为插入失败
这道题要求的是$with positive increments only$,去掉这个限制条件后,以增量序列$1^2, -1^2, 2^2, -2^2 dots, x^2, -x^2$且$x<=p/2$循环试探下一个存储地址即可,证明同上可得(tips:大与2/p的部分可以由p减去小于2/p的部分得到)
还需要注意的一个点:1不是素数,需要在isPrime函数中加以判断
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e6+100;
int p, n, tmp;
bool vis[maxn];
bool isPrime(int x){
if(x==1) return false;
for(int i = 2; i*i <= x; i++)
if(x%i==0) return false;
return true;
}
int main(){
scanf("%d%d", &p, &n);
while(!isPrime(p)) p++;
while(n--){
scanf("%d", &tmp);
int x = tmp%p, y = x, inc = -1;
while(inc<p&&vis[y]) inc++, y = (x+inc*inc)%p;
if(inc!=p) printf("%d", y), vis[y] = 1;
else printf("-");
if(n) printf(" ");
}
}
View Code
补充资料:
线性探测是按线性方法一个一个找,只要表里有空位总能将元素填入;而二次探测有可能出现表中有空间但平方探测找不到的情况
线性探测容易聚集,二次探测聚集情况较线性探测要好。
二次探测有时候探测不到整个散列表空间,是其一大缺陷。但是经过数学家的研究,散列表长度TableSize是某个4k+3(k是正整数)形式的素数时,平方探测法就可以探查到整个散列表空间.
Reference:
https://www.icourse163.org/learn/ZJU-93001?tid=1003997005#/learn/content?type=detail&id=1007588520
https://www.nowcoder.com/discuss/67780
https://en.wikipedia.org/wiki/Quadratic_probing
https://blog.****.net/qq_37142034/article/details/87903983
https://blog.****.net/pennyliang/article/details/5446961
1145 Hashing - Average Search Time (25 分)
这道题相当于1078的扩展,关键在于如何求不在散列表中的元素的平均查找次数。边界值显然为p+1,当查找到已经查找过的单元格后就知道查找失败了;在这个过程中如果发现有空位也能说明该元素不在单元格中
#include <cstdio>
using namespace std;
const int maxn = 1e6+100;
int p, n, m, tmp, h[maxn];
bool vis[maxn];
bool isPrime(int x){
if(x==1) return false;
for(int i = 2; i*i <= x; i++)
if(x%i==0) return false;
return true;
}
int main(){
scanf("%d%d%d", &p, &n, &m);
while(!isPrime(p)) p++;
while(n--){
scanf("%d", &tmp);
int x = tmp%p, y = x, inc = 0;
while(vis[y]&&++inc<p) y = (x+inc*inc)%p;
if(inc!=p) h[y] = tmp, vis[y] = 1;
else printf("%d cannot be inserted.
", tmp);
}
int cnt = m, sum = 0;
while(m--){
scanf("%d", &tmp);
int x = tmp%p, y = x, inc = 0;
while(h[y]!=tmp&&vis[y]&&++inc<p) y = (x+inc*inc)%p;
sum += inc+1;
}
printf("%.1f", 1.0*sum/cnt);
}
View Code
1041 Be Unique (20 分)
水题
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
int n, res, a[maxn];
int t, sta[maxn];
int cnt[maxn];
int main(){
scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
if(cnt[a[i]]==0) sta[++t] = a[i];
cnt[a[i]]++;
}
for(int i = 1; i <= t; i++)
if(cnt[sta[i]]==1) return !printf("%d", sta[i]);
printf("None");
}
View Code
1050 String Subtraction (20 分)
跟那次天梯赛一样,g++不能使用gets(char *),于是换成getline(cin, str)
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
string s1, s2;
bool vis[maxn];
int main(){
getline(cin, s1), getline(cin, s2);
int len1 = s1.length(), len2 = s2.length();
for(int i = 0; i < len2; i++)
if(!vis[s2[i]]) vis[s2[i]] = 1;
for(int i = 0; i < len1; i++)
if(!vis[s1[i]]) printf("%c", s1[i]);
}
//char s1[maxn], s2[maxn];
// gets(s1), gets(s2);
// int len1 = strlen(s1), len2 = strlen(s2);
View Code
Reference:
https://blog.****.net/weixin_41042404/article/details/80934191
1084 Broken Keyboard (20 分)
又是一道水题,不过学到了些英文单词,case insensitive表示不分大小写,capitalized表示大写的
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
string s1, s2;
bool vis[maxn];
int main(){
getline(cin, s1), getline(cin, s2);
int len1 = s1.length(), len2 = s2.length();
for(int i = 0; i < len2; i++) {
if('a'<=s2[i]&&s2[i]<='z') s2[i] += 'A'-'a';
if(vis[s2[i]]) continue;
vis[s2[i]] = 1;
}
for(int i = 0; i < len1; i++){
if('a'<=s1[i]&&s1[i]<='z') s1[i] += 'A'-'a';
if(vis[s1[i]]) continue;
printf("%c", s1[i]), vis[s1[i]] = 1;
}
}
View Code
1092 To Buy or Not to Buy (20 分)
水题+1,注意细节
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
string s1, s2;
int cnt[maxn];
int main(){
getline(cin, s1), getline(cin, s2);
int len1 = s1.length(), len2 = s2.length();
for(int i = 0; i < len2; i++) cnt[s2[i]]++;
for(int i = 0; i < len1; i++){
if(cnt[s1[i]]==0) continue;
cnt[s1[i]]--;
}
int tot = 0;
for(int i = 0; i < len2; i++)
if(cnt[s2[i]]) tot += cnt[s2[i]], cnt[s2[i]] = 0;
if(tot) printf("No %d", tot);
else printf("Yes %d", len1-len2);
}
View Code
大整数运算
1023 Have Fun with Numbers (20 分)
模拟大数相加,一发入魂。不过需要注意的是string未初始化的话是不能对某个下标所在的元素直接赋值的,可以参见这里的讨论:C++中string类字符串可以对其中某个下标元素赋值吗?
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
char s[25], res[25];
int cnt[10];
int main(){
scanf("%s", s);
int len = strlen(s), carry = 0;
for(int i = len-1; i >= 0; i--){
int now = s[i]-'0', tmp = now*2+carry;
tmp > 9 ? carry = 1 : carry = 0;
res[i] = tmp%10+'0';
cnt[now]++, cnt[tmp%10]--;
}
bool flag = true;
for(int i = 0; i <= 9; i++)
if(cnt[i]!=0) {
flag = false;
break;
}
if(flag) printf("Yes
");
else printf("No
");
if(carry) cout << 1;
printf("%s", res);
}
View Code
1024 Palindromic Number (25 分)
仍然是大数模拟,一发就过,关于字符串的翻转可以使用reverse(str.begin(), str.end())和strrev(char *),详细见C++语言中反转字符串的函数strrev(), reverse()
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100;
string s1, s2;
int k;
int main(){
cin >> s1 >> k;
for(int i = 0; i <= k-1; i++){
s2 = s1, reverse(s2.begin(), s2.end());
if(s1==s2) {
cout << s1 << endl << i;
return 0;
}
int len = s1.length(), carry = 0;
for(int i = len-1; i >= 0; i--){
int now = s1[i]-'0'+s2[i]-'0'+carry;
now > 9 ? carry = 1 : carry = 0;
s1[i] = now%10+'0';
}
if(carry==1) s1 = "1" + s1;
}
cout << s1 << endl << k;
}
View Code
思维
1093 Count PAT's (25 分)
利用前缀和计算每个A左右两侧P和T的个数相乘之和即为答案。开始想的是利用P右边A和T的个数来计算,发现这样很麻烦,由于A在中间便换了个计算方式很快就过了。
柳婼的做法是先算出T的个数再去遍历一遍,边计算P和T的个数边计算,这样比我的更加节省空间
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100, mod = 1e9+7;
char s[maxn];
ll t, sta[maxn], cp[maxn], ca[maxn], ct[maxn];
int main(){
scanf("%s", s+1);
int len = strlen(s+1);
for(int i = 1; s[i]; i++){
if(s[i]=='A') sta[++t] = i, ca[i]=ca[i-1]+1, cp[i]=cp[i-1], ct[i]=ct[i-1];
else if(s[i]=='P') cp[i]=cp[i-1]+1, ca[i]=ca[i-1], ct[i]=ct[i-1];
else ct[i]=ct[i-1]+1, ca[i]=ca[i-1], cp[i]=cp[i-1];
}
ll sum = 0;
for(int i = 1; i <= t; i++)
sum = (sum+cp[sta[i]]*(ct[len]-ct[sta[i]]))%mod;
printf("%lld", sum);
}
View Code
1101 Quick Sort (25 分)
distinct positive integers代表不同的正整数,开始没读懂这点导致debug的方向错了,但是要注意这并不是一种排列。由于有这个特性,假如是划分点那么它应该是和排序后的所在位置一样,并且是在包含它的左侧区间中最大的那个数。满足这两个条件便可以是划分点。写完后看了看柳婼的代码,又把自己的代码给优化了下
测试样例2一直格式错误,在网上找了下发现,他要求就算是没有划分点,后面也要输出两个换行,可能是代表空吧
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int maxn = 1e5+100, mod = 1e9+7;
int n, a[maxn], b[maxn], t, sta[maxn], mx[maxn], mn[maxn];
int main(){
scanf("%d", &n);
mx[0] = -1, mn[n+1] = inf;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
for(int i = 1; i <= n; i++)
mx[i] = max(a[i], mx[i-1]), mn[n+1-i] = min(a[n+1-i], mn[n+2-i]);
sort(b+1, b+1+n);
for(int i = 1; i <= n; i++)
if(lower_bound(b+1, b+1+n, a[i])-b==i&&a[i]>mx[i-1]&&a[i]<mn[i+1]) sta[++t] = a[i];
printf("%d
", t);
if(t==0) printf("
");
for(int i = 1; i <= t; i++) {
printf("%d", sta[i]);
if(i!=t) printf(" ");
}
}
优化前
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1e5+100, mod = 1e9+7;
int n, a[maxn], b[maxn], t, sta[maxn], mx;
int main(){
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
sort(b+1, b+1+n);
for(int i = 1; i <= n; i++) {
mx = max(mx, a[i]);
if(b[i]==a[i]&&a[i]==mx) sta[++t] = a[i];
}
printf("%d
", t);
if(t==0) printf("
");
for(int i = 1; i <= t; i++) {
printf("%d", sta[i]);
if(i!=t) printf(" ");
}
}
优化后
Reference:
https://www.liuchuo.net/archives/1917