AtCoderBeginnerContest109题解

第一次AK，真爽qwq

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

*/

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

*/

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

*/

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

*/

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