CF1066D Boxes Packing

来一发通俗易懂的二分题解 直接二分答案 要放$k$个物品就要删掉$n-k$个物品 所以对于一个答案$k$，要放的是$[n-k+1,n]$这$k$个物品 一个一个放，当前盒子能放开就放，放不开就新开一个 ~~感觉代码还是比较简短的~~ ``` #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #define reg register using namespace std; const int N=2e5+5; int n,m,V,a[N]; inline int read() { int x=0,w=1; char c=getchar(); while (!isdigit(c)&&c!='-') c=getchar(); if (c=='-') c=getchar(),w=-1; while (isdigit(c)) { x=(x<<1)+(x<<3)+c-'0'; c=getchar(); } return x*w; } inline bool check(int k) { int res=V,now=1; for (int i=n-k+1;i<=n;i++) { if (res>=a[i]) res-=a[i]; else ++now,res=V-a[i]; } return now<=m; } int main() { n=read(),m=read(),V=read(); for (reg int i=1;i<=n;a[i++]=read()); int l=0,r=n,ans=0; while (l<=r) { int mid=(l+r)>>1; if (check(mid)) ans=mid,l=mid+1; else r=mid-1; } printf("%d\n",ans); return 0; } ```