G - MaratonIME does a competition

水深无声 2022-05-16 09:04 113阅读 0赞

[Statements][]

It's January and MaratonIME is attending to an ACM-ICPC Summer School in Campinas. Renzo, THE POWERFUL, went to visit his students and, as usual, brought chocolates from Peru. However, after a couple of parties, MaratonIME is growing a lot and Renzo doesn't have enough chocolates for everyone.

There is enough chocolate for ![2376bc49fcd2439abc2d7422e7b392c3_v_1534489866][] of the students. So, Renzo will reward the best contestants. As a great coach, Renzo wants MaratonIME to work as a team, so he divided the students in 4 teams, ![90f20285e13f4b1e13d56ff3b7af9b0e_v_1534489866][], ![0a715e90953138d71e8d6a703ebc31aa_v_1534489866][], ![3140dffa42388d0469ce3b3242cc4668_v_1534489866][] and ![ddf1d201664062bbda6798af343ee134_v_1534489866][]. The team which solves more problems wins. If two teams manage to solve the same amount of problems, Renzo rewards the team with smaller lexicographical name (if ![0a715e90953138d71e8d6a703ebc31aa_v_1534489866][] and ![3140dffa42388d0469ce3b3242cc4668_v_1534489866][] draw, ![0a715e90953138d71e8d6a703ebc31aa_v_1534489866][] gets the chocolates).

In order to assign participants to teams, Renzo found out the amount *n* of contestants and gave each one of them a different integer from 1 to *n*. Having numbered the students, Renzo decided that the one with number 1 would go to team ![90f20285e13f4b1e13d56ff3b7af9b0e_v_1534489866][], the one with number 2 would go to team ![0a715e90953138d71e8d6a703ebc31aa_v_1534489866][] and so on. Formally:

*  The contestant with number 1 is on team ![90f20285e13f4b1e13d56ff3b7af9b0e_v_1534489866][]
 *  If 1 < *i* + 1 ≤ *n* and contestant *i* is on ![90f20285e13f4b1e13d56ff3b7af9b0e_v_1534489866][], then contestant *i* + 1 is on team ![0a715e90953138d71e8d6a703ebc31aa_v_1534489866][]
 *  If 1 < *i* + 1 ≤ *n* and contestant *i* is on ![0a715e90953138d71e8d6a703ebc31aa_v_1534489866][], then contestant *i* + 1 is on team ![3140dffa42388d0469ce3b3242cc4668_v_1534489866][]
 *  If 1 < *i* + 1 ≤ *n* and contestant *i* is on ![3140dffa42388d0469ce3b3242cc4668_v_1534489866][], then contestant *i* + 1 is on team ![ddf1d201664062bbda6798af343ee134_v_1534489866][]
 *  If 1 < *i* + 1 ≤ *n* and contestant *i* is on ![ddf1d201664062bbda6798af343ee134_v_1534489866][], then contestant *i* + 1 is on team ![90f20285e13f4b1e13d56ff3b7af9b0e_v_1534489866][]
    
    There are many students and Renzo is busy thinking about the next contest he's creating, so you were chosen to determine which team wins the big prize! You are given a vector *a* of size *n* where, for each index 1 ≤ *i* ≤ *n*, *a**i* is the amount of problems solved by contestant *i* during the contest. The amount of problems solved by a team is the sum of the problems solved by it's contestants.

Input

The first line of input has an integer 4 ≤ *n* ≤ 105. The second line, *n* integers 0 ≤ *a**i* ≤ 104.

Output

You should print one line with name of the team that gets the chocolates: ![90f20285e13f4b1e13d56ff3b7af9b0e_v_1534489866][], ![0a715e90953138d71e8d6a703ebc31aa_v_1534489866][], ![3140dffa42388d0469ce3b3242cc4668_v_1534489866][] or ![ddf1d201664062bbda6798af343ee134_v_1534489866][].

Example

Input

4
    1 2 3 4

Output

题目大意及思路：给你N个数，第一个数去1队，第二个数去2队，第三个数去3队，第4个数去4队，以此类推。最后那个队伍的和最多，输出相对应的ABCD

#include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    int main()
    {
        int n, sum1, sum2, sum3, sum4, i;
        int a[100100];
        int b[101];
        sum1 = sum2 = sum3 = sum4 = 0;
        scanf("%d", &n);
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            if(i % 4 == 1)
            {
                sum1 = sum1 + a[i];
            }
            if(i % 4 == 2)
            {
                sum2 = sum2 + a[i];
            }
            if(i % 4 == 3)
            {
                sum3 = sum3 + a[i];
            }
            if(i % 4 == 0)
            {
                sum4 = sum4 + a[i];
            }
        }
        int max = sum1;
        b[1] = sum1;
        b[2] = sum2;
        b[3] = sum3;
        b[4] = sum4;
        for(i = 1; i <= 4; i++)
        {
            if(b[i] > max)
            {
                max = b[i];
            }
        }
        for(i = 1; i <= 4; i++)
        {
            if(b[i] == max)
            {
                break;
            }
        }
        if(i == 1)
        {
            printf("A\n");
        }
        if(i == 2)
        {
            printf("B\n");
        }
        if(i == 3)
        {
            printf("C\n");
        }
        if(i == 4)
        {
            printf("D\n");
        }
        return 0;
    }

[Statements]: https://odzkskevi.qnssl.com/53a3882263c33a7751e3381d0b176226?v=1534489864
[2376bc49fcd2439abc2d7422e7b392c3_v_1534489866]: https://odzkskevi.qnssl.com/2376bc49fcd2439abc2d7422e7b392c3?v=1534489866
[90f20285e13f4b1e13d56ff3b7af9b0e_v_1534489866]: https://odzkskevi.qnssl.com/90f20285e13f4b1e13d56ff3b7af9b0e?v=1534489866
[0a715e90953138d71e8d6a703ebc31aa_v_1534489866]: https://odzkskevi.qnssl.com/0a715e90953138d71e8d6a703ebc31aa?v=1534489866
[3140dffa42388d0469ce3b3242cc4668_v_1534489866]: https://odzkskevi.qnssl.com/3140dffa42388d0469ce3b3242cc4668?v=1534489866
[ddf1d201664062bbda6798af343ee134_v_1534489866]: https://odzkskevi.qnssl.com/ddf1d201664062bbda6798af343ee134?v=1534489866