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这个题就是大白P144页的一个变形，二分枚举x，对a[i]-x*b[i]从大到小进行排序，选取前n-k个判断和是否大于等于0，若满足条件，增大下限，否则，减小下限。
这个题因为精度问题wa了n次，后来干脆把a[i]和b[i]改成double就过了，循环终止条件写成while (abs(ub - lb) > 1e-4)比写成for(int i=0;i<100;i++)要好，既能减少时间消耗，又能满足精度。
lb和ub初始化时，ub必须等于1.0,否则会莫名其妙的wa
同时注意输出，一开始写成printf("%d ",int(lb*100))wa了n次，正确的应该写成
printf("%.0f ",lb*100);
或者写成
cout << fixed << setprecision(0) << lb * 100 << endl;
AC代码
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <string> #include <vector> #include <deque> #include <list> #include <set> #include <map> #include <stack> #include <queue> #include <numeric> #include <iomanip> #include <bitset> #include <sstream> #include <fstream> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int i=n-1;i>=a;i--) #define in(n) scanf("%d",&(n)) #define in2(x1,x2) scanf("%d%d",&(x1),&(x2)) #define inll(n) scanf("%I64d",&(n)) #define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2)) #define inlld(n) scanf("%lld",&(n)) #define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2)) #define inf(n) scanf("%f",&(n)) #define inf2(x1,x2) scanf("%f%f",&(x1),&(x2)) #define inlf(n) scanf("%lf",&(n)) #define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2)) #define inc(str) scanf("%c",&(str)) #define ins(str) scanf("%s",(str)) #define out(x) printf("%d ",(x)) #define out2(x1,x2) printf("%d %d ",(x1),(x2)) #define outf(x) printf("%f ",(x)) #define outlf(x) printf("%lf ",(x)) #define outlf2(x1,x2) printf("%lf %lf ",(x1),(x2)); #define outll(x) printf("%I64d ",(x)) #define outlld(x) printf("%lld ",(x)) #define outc(str) printf("%c ",(str)) #define pb push_back #define mp make_pair #define fi first #define se second #define SZ(x) ((int)(x).size()) #define mem(X,Y) memset(X,Y,sizeof(X)); typedef vector<int> vec; typedef long long ll; typedef pair<int,int> P; const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1}; const double INF=0x3f3f3f3f; const ll mod=1e9+7; ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} const bool AC=true; int n,k; double a[1005]; //改成int很容易wa double b[1005]; //改成int很容易wa double c[1005]; bool cmp(int x,int y){ return x>y; } bool C(double x){ //传入的为double double sum=0; rep(i,0,n){ c[i]=a[i]-x*b[i]; } sort(c,c+n,cmp); rep(i,0,n-k){ //选取n-k个 sum+=c[i]; } return sum>=0; } int main(){ double lb,mid,ub; while(in2(n,k)==2){ if(n==0&&k==0) break; rep(i,0,n){ inlf(a[i]); } rep(i,0,n){ inlf(b[i]); } lb=0;ub=1;//开的过大可能会因为循环精度不够会莫名奇妙的wa for(int i=0;i<100;i++){ mid=(lb+ub)/2; if(C(mid)) lb=mid; //满足条件,增大下限 else ub=mid; } printf("%.0f ",lb*100);//此处不能写成printf("%d ",int(lb*100)); } return 0; }
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