貪心:Codeforces Round #674 (Div. 3) E Rock, Paper, Scissors s
https://codeforces.com/problemset/problem/1426/E
求A贏的最大次數時很簡單,由於石頭剪刀布贏是一一對應的,直接對應贏法最大情況相加就行。
關鍵是求贏的最小次數,思路是分別求石頭剪刀布能贏的最小次數,然後直接相加即為答案
#include <iostream>
using namespace std;
int main()
{
long long N,ans1=0,ans2=0;
cin>>N;
int b[7];
for(int i=1;i<=6;i++){
cin>>b[i];
}
if(b[1]>b[4]+b[6])ans1+=b[1]-(b[4]+b[6]);
if(b[2]>b[5]+b[4])ans1+=b[2]-(b[5]+b[4]);
if(b[3]>b[6]+b[5])ans1+=b[3]-(b[6]+b[5]);
ans2+=min(b[1],b[5])+min(b[2],b[6])+min(b[3],b[4]);
cout<<ans1<<" "<<ans2<<endl;
return 0;
}
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