test Week2
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2/6 还是欠缺稳定性,心急吃不了热豆腐啊。。keep fighting!!
A 6位的枚举
B 递归枚举,2进制枚举都行
综合两个基础题来看,在紧张的环境下,不能冷静地作出决策,还是多练吧!
C 花的时间不少,一开始的想法是模拟过程,WA, 后来是求区间的方法,也不行,最后还是枚举整个过程最方便,
10000的数据规模,虽然本题求的是时间,但按米为单位是还是可以离散化的。。
a[10000]数组表示a[i]米有车,b[i]同理,那么只要枚举当a[0]~a[8] b[8]~[b16]没有车时,就可以过了
#include <iostream>#include <cstdio>using namespace std;bool a[10005];bool b[10005];int n, m, tmp;int main() {cin >> n;for (int i = 0; i < n; i++) {cin >> tmp;for (int j = tmp; j < tmp + 5; j++) {a[j] = true;}}cin >> m;for (int i = 0; i < m; i++) {cin >> tmp;for (int j = tmp; j < tmp + 5; j++) {b[j] = true;}}double answer = 0;for (int i = 0; i <= 10000; i++) {bool waiting = false;for (int j = i; j < i + 8; j++) if (a[j]) waiting = true;for (int j = i + 8; j < i + 16; j++) if (b[j]) waiting = true;if (!waiting) {answer = i * 3600 / 20000.0; break;}}printf("%.6lf\n", answer);}
C题的区间法终于搞清楚WA在哪了!感谢宏生!当右车道车初始值很小时,可能输出负数!!,怒改!!
#include <iostream>#include <algorithm>#include <cstdio>using namespace std;int car[660];int m , n, tmp;int main () {cin >> n;for (int i = 0; i < n; i++) {cin >> tmp;car[i] = tmp - 8;}cin >> m;for (int i = 0; i < m; i++) {cin >> tmp;car[n+i] = tmp - 16;}int total = n + m;sort(car, car+total);int waiting = 0;if (car[0] >= 0) printf("%f\n", waiting * 0.0);else { waiting = car[0] + 13;for (int i = 1; i < total; i++) {if (car[i] >= waiting) break;else waiting = car[i] + 13;}if (waiting < 0) waiting = 0.0; // key judge!!!!printf("%f\n", waiting * 0.18);}}
D 题实际上依然是枚举检测,它基于这么一个事实:对于两本书的相同数字,先考察其所在的页首个数ab,若相同,或者大于,则不处理,若小于则说明这个数与b-a个数存在错页
#include <iostream>using namespace std;int a, b, n;int main(){cin >> a >> b >> n;long long sum = 0;int topa, topb;for (int i = 0; i < n; i++) {topa = (i / a) * a;topb = (i / b) * b;topb -= topa;if (topb > 0) sum += topb;}cout << sum << endl;}
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