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 + 10, INF = 1e9 + 10, mod = 1e9 + 7; const double eps = 1e-9; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int check(int a, int b, int c) { return a * b * c & 1; } main() { int A = read(), B = read(); if(check(1, A, B) || check(2, A, B) || check(3, A, B)) puts("Yes"); else puts("No"); return 0; } /* 2 2 1 1 1 2 1 1 */
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 + 10, INF = 1e9 + 10, mod = 1e9 + 7; const double eps = 1e-9; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } map<string, bool> mp; int N; string s[MAXN]; main() { int N = read(); for(int i = 1; i <= N; i++) { cin >> s[i]; if(mp.find(s[i]) != mp.end()) {puts("No"); return 0;} if(i > 1) { int l = s[i - 1].length(); if(s[i][0] != s[i - 1][l - 1]) {puts("No"); return 0; } } mp[s[i]] = 1; } puts("Yes"); return 0; } /* 2 2 1 1 1 2 1 1 */
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 + 10, INF = 1e9 + 10, mod = 1e9 + 7; const double eps = 1e-9; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, a[MAXN]; main() { int N = read(), X = read(); for(int i = 1; i <= N; i++) a[i] = read(); int g = abs(X - a[1]); for(int i = 2; i <= N; i++) { int dis = abs(X - a[i]); g = __gcd(g, dis); } printf("%d", g); return 0; } /* 2 2 1 1 1 2 1 1 */
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 = 501, INF = 1e9 + 10, mod = 1e9 + 7; const double eps = 1e-9; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, M; int a[MAXN][MAXN]; int xx[6] = {0, -1, +1, 0, 0}; int yy[6] = {0, 0, 0, -1, +1}; int ans[MAXN * MAXN][7], cnt = 0; void add(int i, int j, int wx, int wy) { ans[++cnt][0] = i; ans[cnt][1] = j; ans[cnt][2] = wx; ans[cnt][3] = wy; } main() { N = read(); M = read(); for(int i = 1; i <= N; i++) for(int j = 1; j <= M; j++) a[i][j] = read(); for(int i = 1; i < N; i++) { for(int j = 1; j <= M; j++) { if(a[i][j] & 1) { int wx = i + 1, wy = j; a[i][j]--; a[wx][wy]++; add(i, j, wx, wy); } } } for(int i = 1; i < M; i++) { if(a[N][i] & 1) { a[N][i]--; a[N][i + 1]++; add(N, i, N, i + 1); } } printf("%d ", cnt); for(int i = 1; i <= cnt; i++) printf("%d %d %d %d ", ans[i][0], ans[i][1], ans[i][2], ans[i][3]); return 0; } /* 2 2 1 1 1 2 1 1 */