HDU 6205 card card card (2017沈阳网赛 - 最大连续子序列和)

题意：

给你两堆数 a 和 b， 每次都是从头开始选择， 选到a-b的和小于0为止， 问你当a 的和最大时，最少操作是多少， 每次操作，可以将头上的a 和b 挪到最后。

思路：

一开始想各种数据结构之类的骚操作。

其实就是一个维护最大连续子序列和。

类似HDU 1003

直接求和a-b  当a-b小于0 时 重新计数， 否则就根据a 的和 来更新答案 和 所在的位置。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1000000 + 10;

int a[maxn << 1];
int b[maxn << 1];

const int inf = 0x3f3f3f3f;
int main(){

int n;

while(~scanf("%d", &n)){
for (int i = 1; i <= n; ++i){
scanf("%d", &a[i]);
a[i + n] = a[i];
}
for (int i = 1; i <= n; ++i){
scanf("%d", &b[i]);
b[i + n] = b[i];
}

int sum = 0;
int suma = 0;
int ans = -inf;
int pos;
int l = 1;
for (int i = 1; i <= 2*n; ++i){
sum += a[i] - b[i];
suma += a[i];
if (suma > ans){
ans = suma;
pos = l;
}

if (sum < 0){
sum = 0;
suma = 0;
l = i + 1;
}
}
printf("%d\n", pos - 1);
}

return 0;
}

card card card

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 877    Accepted Submission(s): 388


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]

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

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