A. Pizza Separation

题解

[codeforce-895A 转载] https://www.cnblogs.com/Roni-i/p/7903036.html
将圆分为连续的两块使其差最小。前缀和或者数组模拟 。

                                        A. Pizza Separation                                        time limit per test1 second                                        memory limit per test256 megabytes                                        inputstandard input                                        outputstandard output

Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut intonpieces. Thei-th piece is a sector of angle equal toai. Vasya and Petya want to divide all pieces of pizza into twocontinuoussectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.

Input
The first line contains one integern(1 ≤n≤ 360) — the number of pieces into which the delivered pizza was cut.

The second line containsnintegersai(1 ≤ai≤ 360) — the angles of the sectors into which the pizza was cut. The sum of allaiis 360.

Output
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.

Examples
input
4
90 90 90 90
output
0
input
3
100 100 160
output
40
input
1
360
output
360
input
4
170 30 150 10
output
0
Note
In first sample Vasya can take1and2pieces, Petya can take3and4pieces. Then the answer is|(90 + 90) - (90 + 90)| = 0.

In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is|360 - 0| = 360.

In fourth sample Vasya can take1and4pieces, then Petya will take2and3pieces. So the answer is|(170 + 10) - (30 + 150)| = 0.

Picture explaning fourth sample:

Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector.

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前缀和：

#include <bits/stdc++.h>using namespace std;int a[400],sum[400];int main(){    int n;    int ans=360;    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%d",&a[i]);        sum[i]=a[i]+sum[i-1];    }    for(int i=0;i<n;i++)    {        for(int j=i;j<n;j++)//t-(360-t)=2*t-360,t为区间所取值，因为是连续区间并且区间可以为空，3个for枚举t或者前缀和或者尺取        {            ans=min(ans,abs(2*(sum[j]-sum[i-1])-360));        }    }    cout<<ans<<endl;    return 0;}

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数组模拟:

#include <bits/stdc++.h>using namespace std;int a[1000],sum;//环的话 数组起码2倍大int main(){    int n;    int ans=360;    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%d",&a[i]);        a[i+n]=a[i];    }    for(int i=0;i<n;i++)    {        sum=0; //注意置0的位置        for(int j=i;j<i+n;j++)        {            sum+=a[j];            ans=min(ans,abs( 2*sum-360) );        }    }    cout<<ans<<endl;    return 0;}/*输入数据直接复制了一遍放到后面,然后枚举拿的起点i，拿的终点j。然后计算拿了多少，差值是多少。*/