AtCoderBeginnerContest109题解

第一次AK，真爽qwq

A

很zz啊，，直接判断三种情况就行

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int check(int a, int b, int c) {
    return a * b * c & ;
}
main() {
    int A = read(), B = read();
    if(check(, A, B) || check(, A, B) || check(, A, B)) puts("Yes");
    else puts("No");

    return ;
}
/*
2 2 1
1 1
2 1 1
*/

A

B

直接模拟即可

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
map<string, bool> mp;
int N;
string s[MAXN];
main() {
    int N = read();

    for(int i = ; i <= N; i++) {
        cin >> s[i];
        if(mp.find(s[i]) != mp.end()) {puts("No"); return ;}
        if(i > ) {
            int l = s[i - ].length();
            if(s[i][] != s[i - ][l - ]) {puts("No"); return ;
            }
        }
        mp[s[i]] = ;
    }
    puts("Yes");

    return ;
}
/*
2 2 1
1 1
2 1 1
*/

B

C

一开始想二分来着，然后发现我zz了。

直接输出 所有距离与起始距离的最大公约数即可

证明显然。。。

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N, a[MAXN];
main() {
    int N = read(), X = read();
    for(int i = ; i <= N; i++) a[i] = read();
    int g = abs(X - a[]);
    for(int i = ; i <= N; i++) {
        int dis = abs(X - a[i]);
        g = __gcd(g, dis);
    }
    printf("%d", g);
    return ;
}
/*
2 2 1
1 1
2 1 1
*/

C

D

一开始读错题了，我以为移动几个都可以，事实上只能移动一个qwq，而且我以为0不统计入答案

我还特地问了一下，真没想到他居然知道我问的什么

考虑一个很显然的正确做法。。

先把所有奇数点往下移动，直到移动到最后一行

再把最后一行的奇数点从左往右移动。。

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N, M;
int a[MAXN][MAXN];
int xx[] = {, -, +, , };
int yy[] = {, , , -, +};
int ans[MAXN * MAXN][], cnt = ;
void add(int i, int j, int wx, int wy) {
    ans[++cnt][] = i;
    ans[cnt][] = j;
    ans[cnt][] = wx;
    ans[cnt][] = wy;
}
main() {
    N = read(); M = read();
    for(int i = ; i <= N; i++)
        for(int j = ; j <= M; j++)
            a[i][j] = read();
    for(int i = ; i < N; i++) {
        for(int j = ; j <= M; j++) {
            if(a[i][j] & ) {
                int wx = i + , wy = j;
                a[i][j]--; a[wx][wy]++;
                add(i, j, wx, wy);
            }
        }
    }
    for(int i = ; i < M; i++) {
        if(a[N][i] & ) {
            a[N][i]--; a[N][i + ]++;
            add(N, i, N, i + );
        }
    }
    printf("%d\n", cnt);
    for(int i = ; i <= cnt; i++)
        printf("%d %d %d %d\n", ans[i][], ans[i][], ans[i][], ans[i][]);
    return ;
}
/*
2 2 1
1 1
2 1 1
*/

D