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