AIM Tech Round 3 (Div. 2)
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AIM Tech Round 3 (Div. 2)
/好慌,补坑。啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊!/
A
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题意:n 个 橘子, 大小大于 b 的不要, 盛橘子汁的容器 为 d, 如果橘子汁溢出了, 那么清理一次容器 使 之内的橘子汁为 0, 问清理多少次。
题解:模拟。注意是大于等于还是大于就可以了,代码很清晰。
代码:
#include <iostream>#include <queue>#include <cstring>#include <cstdio>using namespace std;const int MAXN = 100000 + 10;int ora[MAXN];int main(){ int a, b, d; cin >> a >> b >> d; for(int i = 1; i <= a; i ++) scanf("%d",&ora[i]); int sum = 0, ans = 0; for(int i = 1; i <= a; i ++) { if(ora[i] > b) continue; sum += ora[i]; if(sum > d) { ans ++; sum = 0; } } cout << ans << endl; return 0;}/*Input2 951637 95163844069 951637Output0Answer1*/}
B
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题意:数轴上有 n 个点, 给你所在的位置 a , 求你走过 n-1 个点的最短距离。
题解:分类讨论,弱鸡qer 比较zz。
代码:
#include <iostream>#include <queue>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAXN = 1000000 + 10;int l[MAXN];int getdis(int l, int a, int r){ return min(abs(a-l)*2 + abs(r-a), abs(r-a)*2 + abs(a-l));}int main(){ int n, a; cin >> n >> a; for(int i = 1; i <= n; i ++) scanf("%d",&l[i]); if(n == 1) puts("0"); else{ sort(l+1,l+n+1); int ans = 2147483647; if(a <= l[1]) ans = min(ans, abs(l[n-1] - a)); else if(a >= l[n]) ans = min(ans, abs(l[2] - a)); else if(n == 2) ans = min(abs(a - l[1]), abs(l[n] - a)); else if(l[1] < a && l[2] > a) ans = min(getdis(l[1],a,l[n-1]), abs(l[n] - a)); else if(l[n] > a && l[n-1] < a) ans = min(getdis(l[2],a,l[n]), abs(l[1] - a)); else ans = min(getdis(l[1],a,l[n-1]), getdis(l[2],a,l[n])); printf("%d\n",ans); } return 0; }
终于补上了,嘿嘿嘿、
0 0
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