white 超時 如果 horizon strip numbers 描述 put ont

問題 J: Frosting on the Cake

時間限制: 1 Sec 內存限制: 128 MB
提交: 159 解決: 56
[提交][狀態][討論版][命題人:admin]

題目描述

Iskander the Baker is decorating a huge cake, covering the rectangular surface of the cake with frosting.For this purpose, he mixes frosting sugar with lemon juice and food coloring, in order to produce three kinds of frosting: yellow, pink, and white. These colors are identi?ed by the numbers 0 for yellow,1 for pink, and 2 for white.
To obtain a nice pattern, he partitions the cake surface into vertical stripes of width A1, A2, . . . , An entimeters, and horizontal stripes of height B1, B2, . . . , Bn centimeters, for some positive integer n.
These stripes split the cake surface into n × n rectangles. The intersection of vertical stripe i and horizontal stripe j has color number (i + j) mod 3 for all 1 6 i, j 6 n. To prepare the frosting, Iskander wants to know the total surface in square centimeters to be colored for each of the three colors, and asks for your help.

輸入

The input consists of the following integers:
? on the first line: the integer n,
? on the second line: the values of A1, . . . , An, n integers separated with single spaces,
? on the third line: the values of B1, . . . , Bn, n integers separated with single spaces.
Limits
The input satis?es 3≤n≤100 000 and 1≤A1, . . . , An, B1, . . . , Bn≤10 000.

輸出

The output should consist of three integers separated with single spaces, representing the total area for each color 0, 1, and 2.

樣例輸入

3
1 1 1
1 1 1

樣例輸出

3 3 3
////////////////////////////////////////////////////////////////////////////////////////

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
 
int main()
{
    int a[100005];
    int b[100005
 
];
    int n;
    long long int aa[5]={0};
    long long int bb[5]={0};
    memset(aa,0,sizeof(aa));
    memset(bb,0,sizeof(bb));
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&b[i]);
    }
    for(int i=1;i<=n;i++)
    {
        if(i%3==1)
        {
            aa[1]+=a[i];
            bb[1]+=b[i];
        }
        else if(i%3==2)
        {
            aa[2]+=a[i];
            bb[2]+=b[i];
        }
        else if(i%3==0)
        {
            aa[3]+=a[i];
            bb[3]+=b[i];
        }
    }
//    for(int i=1;i<=3;i++)
//    {
//       printf("%lld %lld\n",aa[i],bb[i]);
//    }
 
long long int ans1,ans2,ans3;
    ans3=aa[1]*bb[1]+aa[3]*bb[2]+aa[2]*bb[3];
    ans1=aa[2]*bb[1]+aa[1]*bb[2]+aa[3]*bb[3];
    ans2=aa[3]*bb[1]+aa[2]*bb[2]+aa[1]*bb[3];
    printf("%lld %lld %lld\n",ans1,ans2,ans3);
}

//如果暴力解會超時 所以找規律 把Ai、Bi分為3部分 然後再計算

問題 J: Frosting on the Cake