Codeforces Round #265 (Div. 2)
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以后做完题及时写blog,再拖拖拉拉剁手!
D. Restore Cube
题意:给8组数据,每组3个数,判断能不能通过改变3个数顺序,使得8组数据构成一个正方体。
解:直接暴力写的,毫无美感。。。还wa了一次。。。3个数有6种组合顺序,深度是8,最多枚举6^8次,大约是10^6,可以水过。关于判断是否构成正方体,对于正方体的每一顶点都满足有3个点到它的距离为a,有3个点到它的距离为sqrt(2)*a,有一个点到它的距离为sqrt(3)*a;
#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,n) for(i=0;i<(n);++i)#define FOR(i,l,h) for(i=(l);i<=(h);++i)#define FORD(i,h,l) for(i=(h);i>=(l);--i)typedef long long LL;const int INF = 0x7fffffff;const double eps = 1e-8;const int N = 1024;const int MOD = 1e9+7;struct node { int x, y, z; node() {}; node(int x, int y, int z) : x(x), y(y), z(z) {} bool operator == (const node& b) { return (this->x == b.x && this->y == b.y && this->z == b.z); }};node nd[8], res[8];double len[8];bool flag;int cube[8][3];double dis(node a, node b) { return double(a.x - b.x)*(a.x - b.x) + double(a.y - b.y)*(a.y - b.y) + double(a.z - b.z)*(a.z - b.z);}bool CK(int k) { int i, j; for(j = 0; j < 8; ++j) { len[j] = dis(nd[k], nd[j]); } sort(len, len+8); for(i = 4; i <= 6; ++i) { if(len[i] != len[1] + len[1]) return false; } if(len[7] != 3*len[1]) return false; return true;}bool check() { for(int i = 0; i < 8; ++i) { if(!CK(i)) return false; } return true;}void dfs(int dep) { if(flag) return ; if(dep == 8) { if(check()) { flag = true; for(int i = 0; i < 8; ++i) { res[i] = nd[i]; } } return ; } int i, j, k; for(i = 0; i < 3; ++i) { for(j = 0; j < 3; ++j) { if(i == j) continue; for(k = 0; k < 3; ++k) { if(k == j || k == i) continue; nd[dep].x = cube[dep][i]; nd[dep].y = cube[dep][j]; nd[dep].z = cube[dep][k]; int xx, yy; for(xx = 0; xx <= dep; ++xx) { for(yy = 0; yy <= dep; ++yy) { if(xx == yy) continue; if(nd[xx] == nd[yy]) break; } if(yy <= dep) break; } if(xx <= dep) continue; dfs(dep + 1); nd[dep].x = nd[dep].y = nd[dep].z = 0; } } }}int main() { //freopen("data.in", "r", stdin); int i, j; for(i = 0; i < 8; ++i) { for(j = 0; j < 3; ++j) { scanf("%d", &cube[i][j]); } } memset(nd, 0, sizeof(nd)); flag = false; dfs(0); if(!flag) puts("NO"); else { puts("YES"); for(i = 0; i < 8; ++i) { printf("%d %d %d\n", res[i].x, res[i].y, res[i].z); } } return 0;}
E. Substitutes in Number
题意:给一个长度为n的数字串,给一组替换规则,替换规则格式为某个数字对应某个数字串,比如1->000,求被所有的替换规则改变以后的数字串对1e9+7取模的结果。
解:利用进制转换的思想,十进制数1024可以表示为1*1000 + 0*100 + 2*10 + 4。暂时称ai*10^i这个柿子中10为进制数,ai为系数。那么,对于某个数字进行替换的时候对应的进制就会有所改变,系数也会有所改变。
比如数据1:
22222->00->70对应的系数变为7,进制为10^0。根据传递性,2对应的系数也变为7,进制数为10^0;
在比如数据2:
12312312->002对应的系数为0,进制数为10^2。
基本思路是在替换之前计算出0-9每个数字替换规则对1e9+7取模后的系数和进制数,然后对原数字串进行替换。
ps:替换规则计算时需要从后往前计算,因为后面的替换会对前面造成影响,比如数据1。
#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,n) for(i=0;i<(n);++i)#define FOR(i,l,h) for(i=(l);i<=(h);++i)#define FORD(i,h,l) for(i=(h);i>=(l);--i)typedef long long LL;const int INF = 0x7fffffff;const double eps = 1e-8;const int N = 1024;const int MOD = 1e9+7;string rp[100010];string S;string rd;string ti;int bi[100010];long long p10[11];long long val[11];int main() { //freopen("data.in", "r", stdin); cin >> S; int n; size_t i; cin >> n; memset(bi, -1, sizeof(bi)); for(int j = 0; j < n; ++j) { cin >> rd; bi[j] = rd[0] - '0'; for(i = 3; i < rd.size(); ++i) { rp[j] += rd[i]; } //cout << rp[bi[j]] << endl; } for(int j = 0; j < 10; ++j) { p10[j] = 10; val[j] = j; } for(int j = n - 1; j >= 0; --j) { long long newp10 = 1; long long newval = 0; int b = bi[j]; for(i = 0; i < rp[j].size(); ++i) { int x = rp[j][i] - '0'; newp10 = (newp10*p10[x]) % MOD; newval = (newval*p10[x] + val[x]) % MOD; } p10[b] = newp10; val[b] = newval; } long long res = 0; for(i = 0; i < S.size(); ++i) { res = (res*p10[S[i] - '0'] + val[S[i] - '0']) % MOD; } cout << res << endl; return 0;}
ps:被完虐啊有木有!好久没刷题了啊有木有!再这么菜b下去就剁手!
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