题目描述
There are n candies in a row, they are numbered from left to right from 1 to n. The size of the i-th candy is ai.
Alice and Bob play an interesting and tasty game: they eat candy. Alice will eat candy from left to right, and Bob — from right to left. The game ends if all the candies are eaten.
The process consists of moves. During a move, the player eats one or more sweets from her/his side (Alice eats from the left, Bob — from the right).
Alice makes the first move. During the first move, she will eat 1 candy (its size is a1). Then, each successive move the players alternate — that is, Bob makes the second move, then Alice, then again Bob and so on.
On each move, a player counts the total size of candies eaten during the current move. Once this number becomes strictly greater than the total size of candies eaten by the other player on their previous move, the current player stops eating and the move ends. In other words, on a move, a player eats the smallest possible number of candies such that the sum of the sizes of candies eaten on this move is strictly greater than the sum of the sizes of candies that the other player ate on the previous move. If there are not enough candies to make a move this way, then the player eats up all the remaining candies and the game ends.
For example, if n=11 and a=[3,1,4,1,5,9,2,6,5,3,5], then:
move 1: Alice eats one candy of size 3 and the sequence of candies becomes [1,4,1,5,9,2,6,5,3,5].
move 2: Alice ate 3 on the previous move, which means Bob must eat 4 or more. Bob eats one candy of size 5 and the sequence of candies becomes [1,4,1,5,9,2,6,5,3].
move 3: Bob ate 5 on the previous move, which means Alice must eat 6 or more. Alice eats three candies with the total size of 1+4+1=6 and the sequence of candies becomes [5,9,2,6,5,3].
move 4: Alice ate 6 on the previous move, which means Bob must eat 7 or more. Bob eats two candies with the total size of 3+5=8 and the sequence of candies becomes [5,9,2,6].
move 5: Bob ate 8 on the previous move, which means Alice must eat 9 or more. Alice eats two candies with the total size of 5+9=14 and the sequence of candies becomes [2,6].
move 6 (the last): Alice ate 14 on the previous move, which means Bob must eat 15 or more. It is impossible, so Bob eats the two remaining candies and the game ends.
Print the number of moves in the game and two numbers:
a — the total size of all sweets eaten by Alice during the game;
b — the total size of all sweets eaten by Bob during the game.
Input
The first line contains an integer t (1≤t≤5000) — the number of test cases in the input. The following are descriptions of the t test cases.
Each test case consists of two lines. The first line contains an integer n (1≤n≤1000) — the number of candies. The second line contains a sequence of integers a1,a2,…,an (1≤ai≤1000) — the sizes of candies in the order they are arranged from left to right.
It is guaranteed that the sum of the values of n for all sets of input data in a test does not exceed 2⋅105.
Output
For each set of input data print three integers — the number of moves in the game and the required values a and b.
Example
input
7
11
3 1 4 1 5 9 2 6 5 3 5
1
1000
3
1 1 1
13
1 2 3 4 5 6 7 8 9 10 11 12 13
2
2 1
6
1 1 1 1 1 1
7
1 1 1 1 1 1 1
output
6 23 21
1 1000 0
2 1 2
6 45 46
2 2 1
3 4 2
4 4 3
题目大意
一共有n堆糖，Alice是从左边开始吃，而Bob从右边开始吃，两个人轮流吃，Alice先开始，下一次吃的人要比上一次吃的人多，最后求出两个人每个人吃了多少颗糖果，以及轮流了多少次。
题目分析
这道题没有什么思维含量，模拟一边过程即可。
a从左往右枚举，b从右往左枚举。再用u存储上次吃的糖数。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=2e5+5;
int c[N]; //存储这n堆糖
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&c[i]);
n--;
int a=0,b=0,u=0,cnt=0; //a存Alice吃的糖数，b存储Bob吃的糖数
//u存储上一次某人吃的糖数，cnt表示轮数。
for(int i=0;i<=n;)
{
cnt++;
int sum=c[i++]; //sum表示这次某人吃的糖数
while(sum<=u&&i<=n) sum+=c[i++]; //算出吃几堆才能大于上次的
a+=sum;
u=sum;
if(i>n) break; //判断糖是否全被a吃完了，是则停止
cnt++;
sum=c[n--]; //因为是从右往左枚举，所以可以通过n--来操作
while(sum<=u&&i<=n) sum+=c[n--];
b+=sum;
u=sum;
}
printf("%d %d %d\n",cnt,a,b); //最后输出即可
}
return 0;
}