- Equation
- Modulo Equality
- Long Beautiful Integer
- Domino for Young
- K Integers
Equation
\[
Time Limit: 3 s\quad Memory Limit: 256 MB
\]
这题做法很多，甚至可以直接暴力判断
view
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
bool ok(int a) {
for(int i=2; i*i<=a; i++) {
if(a%i == 0) return 1;
}
return 0;
}
bool che(int a, int b) {
return ok(a) && ok(b);
}
int main() {
int d;
scanf("%d", &d);
for(int i=1000000000; i>=d; i--) {
if(che(i, i-d)) return 0*printf("%d %d\n", i, i-d);
}
return 0;
}Modulo Equality
\[
Time Limit: 3 s\quad Memory Limit: 256 MB
\]
首先 \(a[1]\) 一定会变成 \(b\) 中的某个元素，那么就可以枚举 \(a[1]\) 变成了多少，把这个数确定为 \(x\)，然后判断合法性并找出所有的 \(x\)。
view
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 2e3 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
int a[maxn], b[maxn];
int s[maxn];
int main() {
scanf("%d%d", &n, &m);
for(int i=1; i<=n; i++) scanf("%d", &a[i]);
for(int i=1; i<=n; i++) scanf("%d", &b[i]);
sort(a+1, a+1+n);
sort(b+1, b+1+n);
int ans = inf;
for(int i=1; i<=n; i++) {
int d = (b[i]-a[1]+m)%m;
for(int j=1; j<=n; j++) s[j] = (a[j]+d)%m;
sort(s+1, s+1+n);
int f = 1;
for(int j=1; j<=n; j++) {
if(s[j] != b[j]) {
f = 0;
break;
}
}
if(f) ans = min(ans, d);
}
printf("%d\n", ans);
return 0;
}Long Beautiful Integer
\[
Time Limit: 3 s\quad Memory Limit: 256 MB
\]
可以发现最后的数字一定是以 \(k\) 为循环节一直循环的，那么我们就可以考虑一开始给出数字的前 \(k\) 位，看用这 \(k\) 位循环能否更大，如果不能的话，把这 \(k\) 位数字加一，然后在开始循环。
由于用 \(k\) 个 \(9\) 来循环一定是可以的，所以不用担心加一后位数变多的问题。
view
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 2e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
char s[maxn], s1[maxn], s2[maxn];
bool ok() {
for(int i=1; i<=n; i++) {
if(s2[i] > s[i]) return true;
if(s2[i] < s[i]) return false;
}
return true;
}
int main() {
scanf("%d%d", &n, &m);
scanf("%s", s+1);
for(int i=1; i<=m; i++) s1[i] = s[i];
for(int i=1, j=1; i<=n; i++) {
s2[i] = s1[j];
j = j%m+1;
}
if(ok()) return 0*printf("%d\n%s\n", n, s2+1);
for(int i=1; i<=m; i++) s1[i] = s1[i]-'0';
s1[m]++;
for(int i=m; i>=1; i--) {
if(s1[i]>=10) {
s1[i] -= 10;
s1[i-1] += 1;
}
s1[i] += '0';
}
for(int i=1, j=1; i<=n; i++) {
s2[i] = s1[j];
j = j%m+1;
}
printf("%d\n%s\n", n, s2+1);
return 0;
}Domino for Young
\[
Time Limit: 3 s\quad Memory Limit: 256 MB
\]
思维题杀我，但是这题的思路真是太优雅了。
我们把整个图看成一个国际棋盘，国际棋盘是黑白相间的，那么也就是说答案一定是 \(min\) (黑格子，白格子)，因为我选了较少的那个，另一个我就一定可以找相邻的凑出来。
view
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
int main() {
scanf("%d", &n);
ll ans1 = 0, ans2 = 0;
for(int i=1, x; i<=n; i++) {
scanf("%d", &x);
if(i&1) ans1+=x/2, ans2+=(x+1)/2;
else ans2+=x/2, ans1+=(x+1)/2;
}
printf("%lld\n", min(ans1, ans2));
return 0;
}K Integers
\[
Time Limit: 3 s\quad Memory Limit: 256 MB
\]
留坑
原文地址：https://www.cnblogs.com/Jiaaaaaaaqi/p/12080426.html
时间： 2024-08-30 17:11:02