T1:YODA
题意：给出两个数，要求把两个数对撞，每一个十进制位上的大的留在原数，如果相同，两个都留在原数，最后输出两个对撞后的原数。 思路：直接模拟即可，两个数每一次都/10%10，比较是否合法。#include#include char s1[20], s2[20];int main() { scanf("%s%s", s1, s2); int N = strlen(s1), M = strlen(s2); for(int i = N-1, j = M-1; i >= 0 && j >= 0; -- i, -- j) { if(s1[i] > s2[j]) s2[j] = 0; else if(s1[i] < s2[j]) s1[i] = 0; } int n = -1, m = -1; for(int i = 0; i < N; ++ i) if(s1[i]) { if(n == -1) n = 0; n = n * 10 + s1[i] - '0'; } for(int i = 0; i < M; ++ i) if(s2[i]) { if(m == -1) m = 0; m = m * 10 + s2[i] - '0'; } if(n == -1) puts("YODA"); else printf("%d\n", n); if(m == -1) puts("YODA"); else printf("%d\n", m); return 0;}
T2:HAN
题意：恶心模拟 思路：步数%26并作出相应操作。#include#include using namespace std;#define LL long longint n, f, cur, st;int cnt[30];char c;inline void GET(int &n) { do c = getchar(); while('0' > c || c > '9'); n = 0; while('0' <= c && c <= '9') {n = n*10+c-'0'; c = getchar();}}int main () { int t1, t2; f = 1; cur = 0; char op[10]; GET(n); for(int i = 1; i <= n; ++ i) { scanf("%s", op); if(op[0] == 'S') { GET(t1); if(t1 - st >= 26) while(cur) { cur = ((cur + f) % 26 + 26) % 26, ++ st; ++ cnt[cur];} for(int i = 0; i < 26; ++ i) cnt[i] += (t1 - st) / 26; st += (t1 - st)/26*26; for(; st < t1; ++ st) { cur = ((cur + f) % 26 + 26) % 26; ++ cnt[cur]; } f *= -1; st = t1; } else { GET(t1); scanf("%s", op); t2 = (op[0] - 'a' + 1)%26; if(t1 - st >= 26) while(cur) { cur = ((cur + f) % 26 + 26) % 26, ++ st; ++ cnt[cur];} for(int i = 0; i < 26; ++ i) cnt[i] += (t1 - st) / 26; st += (t1 - st)/26*26; for(; st < t1; ++ st) { cur = ((cur + f) % 26 + 26) % 26; ++ cnt[cur]; } st = t1; printf("%d\n", cnt[t2]); } } return 0;}
T3:DEATHSTAR
题意：给你一个矩阵 第（i,j）表示Ai&Aj (&表示按位与(bitwise and))求数列{An} 思路：想想或运算，你知道了吗？#includeint a[1005][1005], n;int arr[1005];char c;inline void GET(int &n) { do c = getchar(); while('0' > c || c > '9'); n = 0; while('0' <= c && c <= '9') {n = n*10+c-'0'; c = getchar();}}int main () { GET(n); for(int i = 1; i <= n; ++ i) for(int j = 1; j <= n; ++ j) { GET(a[i][j]); arr[i] |= a[i][j]; arr[j] |= a[i][j]; } printf("%d", arr[1]); for(int i = 2; i <= n; ++ i) printf(" %d", arr[i]); return 0;}
T4:HEWBACCA
题意：给你一个完全K叉树，求两点间的距离。 思路：因为是完全K叉数，那么暴力爬树就是完全没有问题的了（1叉树需要特判）。#include#include using namespace std;#define LL long longlong long p[100], n, m, k, q;char c;inline void GET(LL &n) { do c = getchar(); while('0' > c || c > '9'); n = 0; while('0' <= c && c <= '9') {n = n*10+c-'0'; c = getchar();}}LL fa(LL x) { int t = lower_bound(p+1, p+q+1, x) - p; LL o = x - p[t-1]; o = (o+k-1)/k; return p[t-2]+o;}int dep(LL x) { return lower_bound(p, p+q+1, x) - p; //其实这里可以直接写 "return x;"。因为编号适合深度有关的……我当时SB了}int main () { GET(n); GET(k); GET(m); p[2] = 1; LL t1, t2; if(k != 1) { for(q = 3; n / p[q-1] > k; ++ q) p[q] = p[q-1] * k; for(int i = 1; i <= q; ++ i) p[i] += p[i-1]; p[q] = n; } for(int i = 1; i <= m; ++ i) { int tot = 0; GET(t1); GET(t2); if(t1 > t2) swap(t1, t2); if(k == 1) printf("%I64d\n", t2 - t1); else { while(t1 != t2) { if(dep(t1) < dep(t2)) swap(t1, t2); t1 = fa(t1); ++ tot; } printf("%d\n", tot); } } return 0;}
暂时到这里了，我并不知道T5、T6怎么做……等官方题解吧。