Codeforces Round #609 (Div. 2) 题解

Posted 2021-03-21 jiaaaaaaaqi

• Equation
• Modulo Equality
• Long Beautiful Integer
• Domino for Young
• K Integers

Equation

[ Time Limit: 3 squad 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
", i, i-d);
    }
    return 0;
}

Modulo Equality

[ Time Limit: 3 squad 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
", ans);
    return 0;
}

Long Beautiful Integer

[ Time Limit: 3 squad 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
%s
", 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
%s
", n, s2+1);
    return 0;
}

Domino for Young

[ Time Limit: 3 squad 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
", min(ans1, ans2));
    return 0;
}

K Integers

[ Time Limit: 3 squad Memory Limit: 256 MB ]
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