来自世间,一对平凡的夫妻身旁;来自昨天,谁以青春赴万丈理想?
我是寸土不让的家乡啊,我是永不低头的倔强啊,接过万千热血的初衷,当有对答世界的音量!
A.计算器(calc)
依然不知道为什么会错,但是先按正解改一下好了……
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 105; const ll mod = 1e9 + 7; const int INF = 0x7ffffff; int n, k, cnt[maxn], ma, mb, mc, md; ll Max; inline int read() { int x = 0, f = 1; char ch = getchar(); while(ch > '9' || ch < '0') { if(ch == '-') { f = -1; } ch = getchar(); } while(ch >= '0' && ch <= '9') { x = (x << 1) + (x << 3) + (ch^48); ch = getchar(); } return x * f; } struct node { int id, a, b, c, d; bool operator < (const node &T) const { return id < T.id; } }p[maxn]; ll query(int a, int b, int c, int d) { return (ll)(100+a)*(100+b)*(100+c)*(100+d); } int main() { n = read(); k = read(); for(int i=1; i<=n; i++) { p[i].id = read(); p[i].a = read(); p[i].b = read(); p[i].c = read(); p[i].d = read(); cnt[p[i].id]++; } sort(p+1, p+1+n); for(int i=1; i<=k; i++) { cnt[i] += cnt[i-1]; } for(int i=1; i<=k; i++) { int ha = ma, hb = mb, hc = mc, hd = md; for(int j=cnt[i-1]+1; j<=cnt[i]; j++) { int a = ma+p[j].a, b = mb+p[j].b, c = mc+p[j].c, d = md+p[j].d; ll res = query(a, b, c, d); if(res > Max) { Max = res; ha = a; hb = b; hc = c; hd = d; } } ma = ha; mb = hb; mc = hc; md = hd; } printf("%lld", Max); return 0; }
WA 60eps
感谢能指出以上代码问题的大佬***%%%
B.对称轴(symmetry)
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 2e5 + 3; const ll mod = 1e9 + 7; const int INF = 0x7ffffff; int T, n, ans; double x_1, y_1; inline int read() { int x = 0, f = 1; char ch = getchar(); while(ch > '9' || ch < '0') { if(ch == '-') { f = -1; } ch = getchar(); } while(ch >= '0' && ch <= '9') { x = (x << 1) + (x << 3) + (ch^48); ch = getchar(); } return x * f; } struct node { int x, y; }p[maxn]; double get_k(double x_1, double y_1, double x_2, double y_2) { if(x_1 == x_2) return 1.0*INF; if(y_1 == y_2) return 0.0; return (y_1-y_2) / (x_1-x_2); } bool check(int l, int r, int a, int b, int k) { while(l > a && r > b) { double x_3 = (p[l].x+p[r].x)/2.0, y_3 = (p[l].y+p[r].y)/2.0; double k2 = get_k(x_1, y_1, x_3, y_3); l--; r++; if(abs(k-k2) > 1e-6) return 0; } return 1; } bool check2(int l, int r, int k) { while(l < r) { double x_3 = (p[l].x+p[r].x)/2.0, y_3 = (p[l].y+p[r].y)/2.0; double k2 = get_k(x_1, y_1, x_3, y_3); l++; r--; if(abs(k-k2) > 1e-6) return 0; } return 1; } bool check3(int l, int r, int k) { while(l < r) { double x_3 = (p[l].x+p[r].x)/2.0, y_3 = (p[l].y+p[r].y)/2.0; double k2 = get_k(x_1, y_1, x_3, y_3); l++; r--; if(abs(k-k2) > 1e-6) return 0; } double x_3 = p[l].x, y_3 = p[l].y; double k2 = get_k(x_1, y_1, x_3, y_3); if(abs(k-k2) > 1e-6) return 0; return 1; } double get_len(double x_1, double y_1, double x_2, double y_2) { return (x_1-x_2)*(x_1-x_2)+(y_1-y_2)*(y_1-y_2); } int main() { T = read(); while(T--) { n = read(); ans = 0; for(int i=1; i<=n; i++) { p[i].x = read(); p[i].y = read(); p[n+i].x = p[i].x; p[n+i].y = p[i].y; } if(n & 1) { for(int i=1; i<=n/2; i++) { int j = i+n-1; int lmax = (i+j)>>1, rmin = lmax+1; x_1 = p[i].x, y_1 = p[i].y; int l = lmax, r = rmin; double x_2 = (p[l].x+p[r].x)/2.0, y_2 = (p[l].y+p[r].y)/2.0; if(get_len(x_1, y_1, p[l].x, p[l].y) != get_len(x_1, y_1, p[r].x, p[r].y)) { continue; } double k = get_k(x_1, y_1, x_2, y_2); l--; r++; if(check(l, r, i, i+n, k)) ans++; } } else { for(int i=1; i<=n/2; i++) { int j = i+n-1; int l = i, r = j; x_1 = (p[l].x+p[r].x)/2, y_1 = (p[l].y+p[r].y)/2; l++; r--; double x_2 = (p[l].x+p[r].x)/2.0, y_2 = (p[l].y+p[r].y)/2.0; double k = get_k(x_1, y_1, x_2, y_2); l++; r--;; if(check2(l, r, k)) ans++; } for(int i=1; i<=n/2; i++) { int j = i+n; int l = i, r = j; x_1 = p[l].x, y_1 = p[l].y; l++; r--; double x_2 = (p[l].x+p[r].x)/2.0, y_2 = (p[l].y+p[r].y)/2.0; if(get_len(x_1, y_1, p[l].x, p[l].y) != get_len(x_1, y_1, p[r].x, p[r].y)) { continue; } double k = get_k(x_1, y_1, x_2, y_2); l++; r--; if(check3(l, r, k)) ans++; } } printf("%d/n", ans); } return 0; }
赛时TLE 20eps
就是直接枚举对称轴的直线斜率。。我也不知道我为什么心血来潮的想调它,就调了3个小时。。。
结果就是T3和T4没时间审题了。。。
D.签到题(qiandao)
我没到,,或者说,,我迟到了
未完待续……
原创文章,作者:Maggie-Hunter,如若转载,请注明出处:https://blog.ytso.com/tech/pnotes/277555.html