做了四个题。。
A. Vasya And Password
直接特判即可，，为啥泥萌都说难写，，，，
这个子串实际上是忽悠人的，因为每次改一个字符就可以
/* */ #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 T; char s[MAXN]; void solve() { int N = strlen(s + 1); int a = 0, b = 0, c = 0; for(int i = 1; i <= N; i++) { if(s[i] >= '1' && s[i] <= '9') a++; if(s[i] >= 'a' && s[i] <= 'z') b++; if(s[i] >= 'A' && s[i] <= 'Z') c++; } if(a == 0) { if(b > 1) { for(int i = 1; i <= N; i++) if(s[i] >= 'a' && s[i] <= 'z') {s[i] = '1'; break;} b--; } else if(c > 1) { for(int i = 1; i <= N; i++) if(s[i] >= 'A' && s[i] <= 'Z') {s[i] = '1'; break;} c--; } } if(b == 0) { if(a > 1) { for(int i = 1; i <= N; i++) if(s[i] >= '1' && s[i] <= '9') {s[i] = 'a'; break;} a--; } else if(c > 1) { for(int i = 1; i <= N; i++) if(s[i] >= 'A' && s[i] <= 'Z') {s[i] = 'a'; break;} c--; } } if(c == 0) { if(a > 1) { for(int i = 1; i <= N; i++) if(s[i] >= '1' && s[i] <= '9') {s[i] = 'A'; break;} a--; } else if(b > 1) { for(int i = 1; i <= N; i++) if(s[i] >= 'a' && s[i] <= 'z') {s[i] = 'A'; break;} b--; } } printf("%s\n", s + 1); } main() { T = read(); while(T--) { scanf("%s", s + 1); solve(); } return 0; } /* 2 2 1 1 1 2 1 1 */
B. Relatively Prime Pairs
很显然，$i$和$i+1$是互质的。
做完了
/* */ #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 l, r; main() { l = read(), r = read(); if(l == r) {puts("NO"); return 0;} puts("YES"); for(int i = l; i <= r - 1; i += 2) { cout << i << " " << i + 1 << endl; } return 0; } /* 2 2 1 1 1 2 1 1 */
C. Vasya and Multisets
显然，如果有偶数个优秀的，对半分就可以
如果有奇数个，直接拿出一个$\geqslant 3$的数加到小的里面
否则无解
/* */ #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], timssssss[MAXN]; main() { int n=read(); for(int i = 1; i <= n; ++i) ++timssssss[A[i] = read()]; int cnt[5] = {0,0,0,0,0}; for(int i = 1; i <= 100; ++i) if(timssssss[A[i]] < 3) ++cnt[timssssss[A[i]]]; else ++cnt[3]; if(cnt[1] & 1 && cnt[3] == 0) return puts("NO"),0; puts("YES"); int mid = cnt[1]>>1; if(cnt[1]&1) { int p = 0; for(int i = 1; i <= n; ++i) if(timssssss[A[i]] > 2) { p = i; break; } for(int i = 1 ; i <= n; ++i) if(timssssss[A[i]] == 1) { if(mid) putchar('A'), --mid; else putchar('B'); } else if(i != p) putchar('B'); else putchar('A'); } else { for(int i = 1; i <= n; ++i) if(timssssss[A[i]] == 1) { if(mid) putchar('A'), --mid; else putchar('B'); } else putchar('B'); } return 0; } /* 2 2 1 1 1 2 1 1 */
D. Bicolorings
普及dp？？。。。
因为只有两行，考虑把列的状态记下来
$f[i][j][sta]$表示到第$i$列，有$j$个连通块的方案，当前列的状态为$sta$，就是“白白” “白黑”“黑白”“黑黑”这四种状态
转移的时候枚举上一行选了啥
/* */ #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 = 998244353; 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, K; int f[2001][2001][4]; main() { N = read(); K = read(); f[1][1][0] = 1; f[1][2][1] = 1; f[1][2][2] = 1; f[1][1][3] = 1; for(int i = 2; i <= N; i++) { for(int j = 0; j <= K; j++) { (f[i][j][0] += f[i - 1][j][0]) %= mod; (f[i][j][0] += f[i - 1][j][1]) %= mod; (f[i][j][0] += f[i - 1][j][2]) %= mod; (f[i][j][0] += f[i - 1][j - 1][3]) %= mod; (f[i][j][1] += f[i - 1][j - 1][0]) %= mod; (f[i][j][1] += f[i - 1][j][1]) %= mod; (f[i][j][1] += f[i - 1][j - 2][2]) %= mod; (f[i][j][1] += f[i - 1][j - 1][3]) %= mod; (f[i][j][2] += f[i - 1][j - 1][0]) %= mod; (f[i][j][2] += f[i - 1][j - 2][1]) %= mod; (f[i][j][2] += f[i - 1][j][2]) %= mod; (f[i][j][2] += f[i - 1][j - 1][3]) %= mod; (f[i][j][3] += f[i - 1][j - 1][0]) %= mod; (f[i][j][3] += f[i - 1][j][1]) %= mod; (f[i][j][3] += f[i - 1][j][2]) %= mod; (f[i][j][3] += f[i - 1][j][3]) %= mod; } } int ans = (f[N][K][0] + f[N][K][1] % mod + f[N][K][2] % mod + f[N][K][3] % mod) % mod; cout << ans; return 0; } /* 2 2 1 1 1 2 1 1 */
F. The Shortest Statement