The 2014 ACM-ICPC Asia Mudanjiang Regional Contest - A.Average Score
2014-10-12 13:51
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题目大意:给出两个班级的分数,求平均数的中间范围值。
解题思路:牡丹江现场赛的签到题,分别求出第一个班和第二个班的平均分,然后处理下平均分是不是整数的情况,直接模拟求解即可,详见code。
题目来源:http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5351
code:
解题思路:牡丹江现场赛的签到题,分别求出第一个班和第二个班的平均分,然后处理下平均分是不是整数的情况,直接模拟求解即可,详见code。
题目来源:http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5351
code:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int t,n,m,d; int main(){ // freopen("input.txt","r",stdin); scanf("%d",&t); while(t--){ int ans1=0,ans2=0; scanf("%d%d",&n,&m); for(int i=0;i<n-1;++i){ scanf("%d",&d); ans1+=d; } double maxm=ans1*1.0/(n-1); int tmp=(int)maxm; if((double)tmp==maxm) maxm--; else maxm=tmp; for(int i=0;i<m;++i){ scanf("%d",&d); ans2+=d; } double minm=ans2*1.0/m; tmp=(int)minm; if((double)tmp==minm)minm++; else minm=tmp+1; printf("%d %d\n",(int)minm,(int)maxm); } return 0; }
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