摘要：
题目链接emmmm...思路是群里群巨聊天讲这题是用尺取法.....emmm然后就没难度了，不过时间上3000多，有点.....盗了个低配本的读入挂发现就降到2800左右，翻了下，发现神犇Claris280MS秒过.......%%%#include#include#include#include#include˂iostre
题目链接
emmmm...思路是群里群巨聊天讲这题是用尺取法.....emmm然后就没难度了，不过时间上3000多，有点.....盗了个低配本的读入挂发现就降到2800左右，
翻了下，发现神犇Claris280MS秒过.......%%%
#include <stdio.h>#include <stdlib.h>#include <cmath>#include <string.h>#include <iostream>#include <algorithm>#include <queue>#include <vector>#include <string>#include <ctype.h> //****************************************************** #define lrt (rt*2) #define rrt (rt*2+1) #define LL long long #define inf 0x3f3f3f3f #define pi acos(-1.0) #define exp 1e-8 //*************************************************** //*********************************** #define eps 1e-8 #define inf 0x3f3f3f3f #define INF 2e18 #define LL long long #define ULL unsigned long long #define PI acos(-1.0) #define pb push_back #define mk make_pair #define all(a) a.begin(),a.end() #define rall(a) a.rbegin(),a.rend() #define SQR(a) ((a)*(a)) #define Unique(a) sort(all(a)),a.erase(unique(all(a)),a.end()) #define min3(a,b,c) min(a,min(b,c)) #define max3(a,b,c) max(a,max(b,c)) #define min4(a,b,c,d) min(min(a,b),min(c,d)) #define max4(a,b,c,d) max(max(a,b),max(c,d)) #define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e)) #define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e)) #define Iterator(a) __typeof__(a.begin()) #define rIterator(a) __typeof__(a.rbegin()) //#define FastRead ios_base::sync_with_stdio(0);cin.tie(0) #define FastRead ios::sync_with_stdio(false); #define CasePrint pc('C'); pc('a'); pc('s'); pc('e'); pc(' '); write(qq++,false); pc(':'); pc(' ') #define vi vector <int> #define vL vector <LL> #define For(I,A,B) for(int I = (A); I < (B); ++I) #define FOR(I,A,B) for(int I = (A); I <= (B); ++I) #define rFor(I,A,B) for(int I = (A); I >= (B); --I) #define Rep(I,N) For(I,0,N) #define REP(I,N) FOR(I,1,N) using namespacestd; const int maxn=2e6+50; intn,a[maxn],b[maxn]; intmain() { while(~scanf("%d",&n)) { REP(i,n) scanf("%d",&a[i]),a[i+n]=a[i]; REP(i,n) scanf("%d",&b[i]),b[i+n]=b[i]; int sum=a[1]-b[1],f=1; FOR(i,2,2*n) { sum+=(a[i]-b[i]); if(i-f+1 == n) break; if(sum<0) f=i+1,sum=0; } printf("%d ",f-1); } return 0; }