[2911] Luo Gu cattle bones

Title Description

Bessie loves board games and role-playing games so she persuaded Farmer John to drive her to the hobby shop where she purchased three dice for rolling. These fair dice have S1, S2, and S3 sides

respectively (2 <= S1 <= 20; 2 <= S2 <= 20; 2 <= S3 <= 40).

Bessie rolls and rolls and rolls trying to figure out which three-dice sum appears most often.

Given the number of sides on each of the three dice, determine which three-dice sum appears most frequently. If more than one sum can appear most frequently, report the smallest such sum.

POINTS: 70

Input Format

* Line 1: Three space-separated integers: S1, S2, and S3

Output Format

* Line 1: The smallest integer sum that appears most frequently when the dice are rolled in every possible combination.

Translation of the meaning of problems

Bessie like to play board games and role-playing games, so she persuaded John drove her to a small shop where she bought three dice. The three face different number of dice are s1 s2 s3

For numbers on each side of a dice surface S is 1, 2, ..., S probability. (Digital on) each face appear equal. Bessie want to find out all the "three digital and face "in which the probability of occurrence of the maximum value and.

Now given the number of sides of each of the dice, which all need to find the "digital three surfaces and" appears most frequently. If there are many probabilities and appeared the same, you only need a minimum output of the data range : (2 <S1 <= 20; 2 <= S2 <= 20; 2 <= S3 <= 40)

Sample input and output

Input # 1copy

3 2 3 

Output # 1copy

5 

Description / Tips

Here are all the possible outcomes.

1 1 1 -> 3  
1 2 1 -> 4  
2 1 1 -> 4  
2 2 1 -> 5  
3 1 1 -> 5  
3 2 1 -> 6 
1 1 2 -> 4  
1 2 2 -> 5  
2 1 2 -> 5  
2 2 2 -> 6  
3 1 2 -> 6  
3 2 2 -> 7 
1 1 3 -> 5  
1 2 3 -> 6  
2 1 3 -> 6  
2 2 3 -> 7  
3 1 3 -> 7  
3 2 3 -> 8

Both 5 and 6 appear most frequently (five times each), so 5 is the answer.

Solution: expect a title problem? ? Night scholarship ah. Hey!

           (Although the title is red but still very shameless paid up)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
typedef long long ll;
using namespace std;
int rt,rp,ic;
int main(){
    scanf("%d %d %d", &rt, &rp, &ic);
    if(ic>rt+rp) printf("%d", rt+rp+1);
    else if(rp>rt+ic) printf("%d", rt+ic+1);
    else if(rt>ic+rp) printf("%d", ic+rp+1);
    else printf("%d", (rt+rp+ic+3)/2);
    return 0;
}