Codeforces Round #317 (Div. 2) (572A Arrays,572B Order Book)
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1.572A Arrays
题目链接:
http://codeforces.com/problemset/problem/572/A
解题思路:
取第一个数列的第k个数,然后倒着取第二个数列的第m个数,比较即可。。。
AC代码:
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;int a[100005];int b[100005];int main(){ int l1,l2; while(~scanf("%d%d",&l1,&l2)){ int k,m; scanf("%d%d",&k,&m); for(int i = 1; i <= l1; i++) scanf("%d",&a[i]); for(int i = 1; i <= l2; i++) scanf("%d",&b[i]); if(a[k] < b[l2-m+1]) printf("YES\n"); else printf("NO\n"); } return 0;}
2.572B Order Book
题目链接:
http://codeforces.com/problemset/problem/572/B
解题思路:
英语不好的人,读题目起来真费解,特别是大半夜做题。。。那滋味。。。
题目大意:
给你n个做账记录(乱序给出),如果相同类且价格一样,则数量合并,取sell价格从小到大前s个,再取buy价格从大到小s个,最后先将sell价格从大到小输出,再将buy价格从大到小输出。。。
模拟即可。。。
AC代码:
#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;const int maxn = 100005;int buy[maxn],sell[maxn];struct node{ int p,q;};vector<node> vb,vs;int main(){ int n,s; while(~scanf("%d%d",&n,&s)){ char op[5]; int p,q; memset(buy,0,sizeof(buy)); memset(sell,0,sizeof(sell)); for(int i = 0; i < n; i++){ scanf("%s%d%d",op,&p,&q); if(op[0] == 'B') buy[p] += q; else sell[p] += q; } vb.clear(); vs.clear(); for(int i = 0; i < maxn && vs.size() < s; i++) if(sell[i]) vs.push_back(node{i,sell[i]}); for(int i = maxn; i >= 0 && vb.size() < s; i--) if(buy[i]) vb.push_back(node{i,buy[i]}); for(int i = vs.size()-1; i >= 0; i--) printf("S %d %d\n", vs[i].p, vs[i].q); for (int i = 0; i < vb.size(); i++) printf("B %d %d\n", vb[i].p, vb[i].q); } return 0;}
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