# HDU 6205 （模拟） card card card

2018/03/12 08:14

Problem Description As a fan of Doudizhu, WYJ likes collecting playing cards very much. One day, MJF takes a stack of cards and talks to him: let's play a game and if you win, you can get all these cards. MJF randomly assigns these cards into n heaps, arranges in a row, and sets a value on each heap, which is called "penalty value". Before the game starts, WYJ can move the foremost heap to the end any times. After that, WYJ takes the heap of cards one by one, each time he needs to move all cards of the current heap to his hands and face them up, then he turns over some cards and the number of cards he turned is equal to the penaltyvalue. If at one moment, the number of cards he holds which are face-up is less than the penaltyvalue, then the game ends. And WYJ can get all the cards in his hands (both face-up and face-down). Your task is to help WYJ maximize the number of cards he can get in the end.So he needs to decide how many heaps that he should move to the end before the game starts. Can you help him find the answer? MJF also guarantees that the sum of all "penalty value" is exactly equal to the number of all cards.

Input There are about 10 test cases ending up with EOF. For each test case: the first line is an integer n (1≤n≤106), denoting n heaps of cards; next line contains n integers, the ith integer ai (0≤ai≤1000) denoting there are ai cards in ith heap; then the third line also contains n integers, the ith integer bi (1≤bi≤1000) denoting the "penalty value" of ith heap is bi.

Output For each test case, print only an integer, denoting the number of piles WYJ needs to move before the game starts. If there are multiple solutions, print the smallest one.

Sample Input 5 4 6 2 8 4 1 5 7 9 2

Sample Output 4 Hint

[pre] For the sample input:

• If WYJ doesn't move the cards pile, when the game starts the state of cards is: 4 6 2 8 4 1 5 7 9 2 WYJ can take the first three piles of cards, and during the process, the number of face-up cards is 4-1+6-5+2-7. Then he can't pay the the "penalty value" of the third pile, the game ends. WYJ will get 12 cards.
• If WYJ move the first four piles of cards to the end, when the game starts the state of cards is: 4 4 6 2 8 2 1 5 7 9 WYJ can take all the five piles of cards, and during the process, the number of face-up cards is 4-2+4-1+6-5+2-7+8-9. Then he takes all cards, the game ends. WYJ will get 24 cards.

It can be improved that the answer is 4.

huge input, please use fastIO. [/pre]

5

2 8 4 4 6

7 9 2 1 5

2-7 拿到第1堆的时候游戏就结束，获得排数：2

8 4 4 6 2

9 2 1 5 7

8-9 = -1 拿到第一堆牌的时候游戏结束，获得牌数： 8

4 4 6 2 8

2 1 5 7 9

4-2 + 4-1 + 6-5 + 2- 7 + 8-9 > 0.则牌全拿完了。

``````#include <iostream>
#include <stdio.h>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1e6;
int num[maxn];//  牌数
int value[maxn];  //权值
int a[maxn];  //差值
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i = 1; i <= n; i++)
scanf("%d",&num[i]);
for(int i = 1; i <= n; i++)
{
scanf("%d",&value[i]);
a[i] = num[i]-value[i];
}
int Max = -1;   ///获得的最大卡片数
int total_cards = 0; ///当前总卡片数
int sum = 0;         ///求和
int ans = 0;         ///答案
int start = 1;       ///第几堆牌是开始
for(int i = start; i <= n; i++)
{
sum += a[i];
total_cards += num[i];
if(sum < 0)      ///游戏结束
{
if(total_cards > Max)  ///更新答案
{
Max = total_cards;
ans = start;
}
start = i+1;  ///下一次开始的位置是i+1
sum = 0;
total_cards = 0;
}
}
///从start ~ n 还要再加上从1~start，因为他们开始前被挪到了后边
for(int i = 1; i < start; i++)
{
sum += a[i];
total_cards += num[i];
if(sum < 0)
{
if(total_cards > Max)
{
Max = total_cards;
ans = start;
}
break;
}
}
if(total_cards > Max)
{
Max = total_cards;
ans = start;
}
printf("%d\n",ans-1);  ///从第ans开始，执行操作ans-1次
}
return 0;
}

``````

0
0 收藏

0 评论
0 收藏
0