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HDU - 6011 Lotus and Characters
仔细读题,然后来一发大模拟!
Lotus has nn kinds of characters,each kind of characters has a value and a amount.She wants to construct a string using some of these characters.Define the value of a string is:its first character‘s value*1+its second character‘s value *2+...She wants to calculate the maximum value of string she can construct.
Since it‘s valid to construct an empty string,the answer is always ≥0≥0。
Since it‘s valid to construct an empty string,the answer is always ≥0≥0。
InputFirst line is T(0≤T≤1000)T(0≤T≤1000) denoting the number of test cases.
For each test case,first line is an integer n(1≤n≤26)n(1≤n≤26),followed by nn lines each containing 2 integers vali,cnti(|vali|,cnti≤100)vali,cnti(|vali|,cnti≤100),denoting the value and the amount of the ith character.
OutputFor each test case.output one line containing a single integer,denoting the answer.
Sample Input
2 2 5 1 6 2 3 -5 3 2 1 1 1
Sample Output
35 5
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <vector> using namespace std; int main() { int T; scanf("%d",&T); while(T--) { vector<int> vec; int n; scanf("%d",&n); int allnum=0; for(int i=0;i<n;i++) { int val,num; scanf("%d%d",&val,&num); allnum+=num; while(num--) { vec.push_back(val); } } //代码里最重要的是dissum这个数,代表着ipos位置后面的数字各取一个的和 sort(vec.begin(),vec.end()); int ipos=0; long long int disnum=0; //int flag=1; for(int i=0;i<allnum;i++) { if(vec[i]<0) { ipos=max(ipos,i); } else if(vec[i]>=0) { disnum+=vec[i]; } } //cout<<ipos<<endl; long long int ans=0; if(ipos!=0) {for(int i=ipos+1;i<allnum;i++) { ans+=vec[i]*(i-ipos); } while((vec[ipos]+disnum)>0&&ipos>=0) { ans+=(vec[ipos]+disnum); disnum+=vec[ipos];//dissum是会改变的ipos位置变化会引起这个的变化 ipos--; }} else for(int i=0;i<allnum;i++) { ans+=vec[i]*(i+1); } //cout<<ans<<endl; printf("%lld\n",ans); } }
HDU - 6011 Lotus and Characters
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