Codeforces Round #228 (Div. 2) 题解
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A.Fox and Number Game
题意:有一列数a(100),可以进行多次操作,每次选出数组中的两个数xi>xj,然后xi = xi – xj,问这n个数的和最小是多少
思路:数组中的数最后都会变成这n个数的gcd,所以求出数组的gcd,然后乘以数组个数就是最后的答案
代码:
#include <bits/stdc++.h> using namespace std; int main() { int n; scanf("%d", &n); int ans; scanf("%d", &ans); for(int i = 2; i <= n; i++) { int x; scanf("%d", &x); ans = __gcd(ans, x); } printf("%d ",ans * n); return 0; }
B. Fox and Cross
题意:给一张只含有’.’,’#’,的地图,问这种地图中的#号是不是由
.#.
###
.#.
这样的十字构成的
思路:遇到的每一个十字都填上,贪心的填,最后检查一下是否还有#号
代码:
#include <bits/stdc++.h> using namespace std; char mp[105][105]; int main() { int n; scanf("%d", &n); for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { cin>>mp[i][j]; } } for(int i = 2; i < n; i++) { for(int j = 2; j < n; j++) { if(mp[i][j] == ‘#‘ && mp[i - 1][j] == ‘#‘ && mp[i + 1][j] == ‘#‘ && mp[i][j - 1] == ‘#‘ && mp[i][j + 1] == ‘#‘) { mp[i][j] = ‘.‘; mp[i + 1][j] = ‘.‘; mp[i - 1][j] = ‘.‘; mp[i][j + 1] = ‘.‘; mp[i][j - 1] = ‘.‘; } } } bool ok = true; for(int i = 1; i <= n && ok; i++) { for(int j = 1; j <= n && ok; j++) { if(mp[i][j] == ‘#‘) ok = false; } } if(ok) puts("YES"); else puts("NO"); return 0; }
C. Fox and Box Accumulation
题意:有n个箱子,每个箱子有一个ai,表示这个箱子上面最多还能放几个箱子,问这n个箱子最少放城几堆
思路:二分最后放成几堆,对于mid堆,先把最大的mid的放上去,然后每次维护这一堆还能放几个
代码:
#include <bits/stdc++.h> using namespace std; const int maxn = 107; int n; int a[maxn], c[maxn]; bool cmp(int x, int y) { return x > y; } bool check(int x) { for(int i = 1; i <= x; i++) c[i] = a[i]; int j = 1; for(int i = x + 1; i <= n; i++) { bool ok = true; int pre = j; for(;j <= x;) { // printf("test i==%d j==%d c[j]==%d ",i,j,c[j]); if(c[j] > 0) { c[j] = min(c[j] - 1, a[i]); ok = false; j++; if(j == x + 1) j = 1; break; } j++; if(j == x + 1) j = 1; if(j == pre) break; } if(ok) return false; } return true; } int main() { scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); } sort(a + 1, a + 1 + n, cmp); int L = 1, R = n; int ans = 0; // printf("test %d ",check(1)); while(L <= R) { int mid = (L + R) >> 1; // printf("test %d %d %d ", L, R, mid); if(check(mid)) { R = mid - 1; ans = mid; } else L = mid + 1; } printf("%d ", ans); return 0; }
D. Fox and Minimal path
题意:最多用1000个点,构造一个从1到2有k(1e9)条最短路的图
思路:把k进行二进制拆分,然后先把最大的2^x次方的路画出来,然后在画一条等长的路,把中间的2^x次方连起来
代码:
#include <bits/stdc++.h> using namespace std; int mp[1007][1007]; int a[57]; int main() { int k; scanf("%d", &k); int cnt = 0; while(k) { a[cnt++] = k & 1; k >>= 1; } int now = 3; mp[1][3] = mp[3][1] = 1; for(int i = 0; i < cnt - 1; i++) { mp[now][now + 1] = 1; mp[now + 1][now] = 1; mp[now][now + 2] = 1; mp[now + 2][now] = 1; mp[now + 1][now + 3] = 1; mp[now + 3][now + 1] = 1; mp[now + 2][now + 3] = 1; mp[now + 3][now + 2] = 1; now += 3; } int res = now + 1; mp[2][now] = mp[now][2] = 1; for(int i = 0; i < (cnt - 2) * 2; i++, res++) { mp[res][res + 1] = 1; mp[res + 1][res] = 1; } mp[res][now] = mp[now][res] = 1; int pos = now + 1; now = res; int as = 1; for(int i = 0; i < cnt; i++) { if(a[i]) { mp[as * 3][pos] = 1; mp[pos][as * 3] = 1; } as ++; pos += 2; } printf("%d ",now); for(int i = 1; i <= now; i++) { for(int j = 1; j <= now; j++) { if(mp[i][j])printf("Y"); else printf("N"); } puts(""); } return 0; }
E. Fox and Card Game
题意:有n(100)堆牌,每堆的牌的数量为s(100),每张牌上都有一个分值,A每次只能从一堆的最上面拿,B只能从一堆的最下面拿,A先手,AB都是绝顶聪明的,问最后AB各拿了多少分
思路:因为AB都是决定聪明的,所以对于一堆偶数的牌,每个人都拿一半,然后只剩下奇数的牌,对于中间的那张牌排序,轮流那就可以
代码:
#include <bits/stdc++.h> using namespace std; const int maxn = 105; int a[maxn]; int stk[maxn], top; int n; bool cmp(int x, int y) { return x > y; } int main() { scanf("%d", &n); int ansx = 0, ansy = 0; for(int i = 1; i <= n; i++) { int op; scanf("%d", &op); for(int i = 1; i <= op; i++) { scanf("%d", &a[i]); } int up = op / 2; for(int j = 1; j <= up; j++) { ansx += a[j]; ansy += a[op - j + 1]; } if(op & 1) stk[++top] = a[up + 1]; } sort(stk + 1, stk + 1 + n, cmp); for(int i = 1; i <= top; i++) { if(i & 1) ansx += stk[i]; else ansy += stk[i]; } printf("%d %d ", ansx, ansy); return 0; }
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