HDU-4974-a simple water problem (Greedy + evidentiary)

Question: n teams (n <= 100000), each team has a total score of AI (AI <= 1000000). Each team can score one point at most for each game, ask the minimum number of matches.

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4974

--> We should try our best to set the score of each game to 1: 1, so that we can achieve the minimum number of matches (not less than the score of a single team ).

Assume that the score for two matches is 1: 0,

1) A: B = 1: 0, C: D = 1: 0. In this case, you can arrange a: c = 1: 1. the same score can be obtained in only one field.

2) A: B = 1: 0, A: c = 1: 0, take another game D: E = 1: 1, then you can arrange a: D = 1: 1, A: E = 1: 1. the same score can be obtained in only two places.

Therefore, the score for one game is 1: 0 at most, and for other games, the score is 1: 1. Therefore, the result is = max (the highest score for a single team, (All scores and + 1)/2 )... (Note range: 10 ^ 5*10 ^ 6> 2 ^ 31-1)

For submission on virtual contest, you must enable and disable the input to avoid TLE ..

Scanf in HDU library 4974 can be used as an AC ..

#include <cstdio>int ReadInt(){    int ret = 0;    char ch;    while ((ch = getchar()) && ch >= '0' && ch <= '9')    {        ret = ret * 10 + ch - '0';    }    return ret;}int main(){    int T, N, a, kase = 0;    scanf("%d", &T);    getchar();    while (T--)    {        long long sum = 0;        long long ret = 0;        N = ReadInt();        while (N--)        {            a = ReadInt();            sum += a;            if (a > ret)            {                ret = a;            }        }        if (sum & 1)        {            sum++;        }        sum >>= 1;        if (sum > ret)        {            ret = sum;        }        printf("Case #%d: %I64d\n", ++kase, ret);    }    return 0;}

HDU-4974-a simple water problem (Greedy + evidentiary)