#1831 : 80 Days #1831 : 80 Days 時間限制:1000ms 單點時限:1000ms 記憶體限制:256MB 描述 80 Days is an interesting game based on Jules Verne's science fiction "Around th ...
#1831 : 80 Days
時間限制:1000ms 單點時限:1000ms 記憶體限制:256MB描述
80 Days is an interesting game based on Jules Verne's science fiction "Around the World in Eighty Days". In this game, you have to manage the limited money and time.
Now we simplified the game as below:
There are n cities on a circle around the world which are numbered from 1 to n by their order on the circle. When you reach the city i at the first time, you will get ai dollars (ai can even be negative), and if you want to go to the next city on the circle, you should pay bi dollars. At the beginning you have c dollars.
The goal of this game is to choose a city as start point, then go along the circle and visit all the city once, and finally return to the start point. During the trip, the money you have must be no less than zero.
Here comes a question: to complete the trip, which city will you choose to be the start city?
If there are multiple answers, please output the one with the smallest number.
輸入
The first line of the input is an integer T (T ≤ 100), the number of test cases.
For each test case, the first line contains two integers n and c (1 ≤ n ≤ 106, 0 ≤ c ≤ 109). The second line contains n integers a1, …, an (-109 ≤ ai ≤ 109), and the third line contains n integers b1, …, bn (0 ≤ bi ≤ 109).
It's guaranteed that the sum of n of all test cases is less than 106
輸出
For each test case, output the start city you should choose.
提示
For test case 1, both city 2 and 3 could be chosen as start point, 2 has smaller number. But if you start at city 1, you can't go anywhere.
For test case 2, start from which city seems doesn't matter, you just don't have enough money to complete a trip.
- 樣例輸入
-
2 3 0 3 4 5 5 4 3 3 100 -3 -4 -5 30 40 50
- 樣例輸出
-
2 -1
比賽的時候都以為是dp，但是自己寫的感覺就是暴力。。。
思路：p[i]=a[i]-b[i] 化環為直 尺取法取長度為n的序列，front如果大於n就是-1 （見註釋）1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <string.h> 5 #include <stdlib.h> 6 #include <string> 7 #include <vector> 8 #include <set> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 #include <sstream> 13 #include <stack> 14 using namespace std; 15 #define rep(i,a,n) for (int i=a;i<n;i++) 16 #define per(i,a,n) for (int i=n-1;i>=a;i--) 17 #define pb push_back 18 #define mp make_pair 19 #define all(x) (x).begin(),(x).end() 20 #define fi first 21 #define se second 22 #define SZ(x) ((int)(x).size()) 23 #define FO freopen("in.txt", "r", stdin); 24 #define debug(x) cout << "&&" << x << "&&" << endl; 25 #define lowbit(x) (x&-x) 26 #define mem(a,b) memset(a, b, sizeof(a)); 27 typedef vector<int> VI; 28 typedef long long ll; 29 typedef pair<int,int> PII; 30 const ll mod=1000000007; 31 const int inf = 0x3f3f3f3f; 32 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;} 33 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 34 //head 35 36 const int N=1e6+5; 37 int p[N<<1],a[N],b[N],_,n; 38 ll c; 39 int main() { 40 for(scanf("%d",&_);_;_--) { 41 scanf("%d%lld",&n,&c); 42 rep(i,1,n+1) scanf("%d",&a[i]); 43 rep(i,1,n+1) scanf("%d",&b[i]); 44 rep(i,1,n+1) p[i]=p[i+n]=a[i]-b[i];//模擬環 45 int front=1,rear=1;//都指向1 46 while(front<=n&&rear-front+1<=n) {//front<=n n長度 47 c+=p[rear];//累加 後移 48 rear++; 49 while(c<0){//如果c<0 說明front不能到達rear front後移 50 c-=p[front]; 51 front++; 52 } 53 } 54 if(front>n) puts("-1");//1~n都不行 55 else printf("%d\n",front);//可以取長度為n的序列 輸出隊頭 56 } 57 }
（貌似提交是過不去的。。。）
