二分答案即可，check方法很简单，每个元素比预期少总和的和多出来的的综合是不是两倍关系即可 #include<bits/stdc++.h> using namespace std; int main(){ long long a[4]; long long sum = 0; for(int i=0;i<4;i++){ scanf("%lld",&a[i]); sum += a[i]; } sort(a,a+4); long long l = 1; long long r = sum / 4; long long ans = -1; while(l<=r){ long long mid = (l+r) / 2; long long need = 0; long long left = 0; for(int i=0;i<4;i++){ if(a[i] < mid){ need += (mid - a[i]); }else if(a[i]>mid){ left += (a[i] - mid); } } if(need * 2<=left){ ans = max(ans,mid); l = mid+1; }else{ r = mid-1; } } cout<<ans * 4<<endl; }