## Elections

Time limit: 2000 ms

Memory limit: 65536 KB

### Description

There are two candidates campaigning to be president of a country. From newspaper polls, it is clear what percentages of people plan to vote for each candidate in each state. Candidate 1 wants to campaign in one last state, and needs to figure out which state that should be.

Given some states and votes in each state, find one city with lowest percentage of people planning on voting for candidate 1.

### Input Format

First line consists of an integer N (1 <= N <=50), the number of states. Each of the next N lines is a string (1 to 50 characters long, inclusive) consisting of the characters ‘1’ and ‘2’, where ‘1’ represents some number of votes for candidate 1, and ‘2’ represents votes for candidate 2 (in each state, every character represents the same number of votes).

### Output Format

An integer, the 0-based index of the state where the lowest percentage of people are planning on voting for candidate 1 (lowest percentage of ‘1’ characters in that element of the input). If there are multiple such states, return one with the lowest index

```3
1222
1122
1222
```

`0`

### Note

In the first state only 25% of people prefer candidate 1, while in the second and third, 50% and 25% prefer him, respectively.

### Sample Input 2

```3
1222111122
2222222111
11111222221222222222
```

`1`

### Note

The percentages of people, prefering candidate 1 to candidate 2 are (in order): 50%, 30%, 30%

```8
111
112
121
122
211
212
221
222
```

`7`

```6
1122
1221
1212
2112
2121
2211
```

`0`

```6
11112222111121
1211221212121
112111222
11122222222111
112121222
1212122211112
```

### Sample Output 5

`3`

Solution

```#include <stdio.h>
#include <stdlib.h>

int main()
{
int N,i,j;
double tempVal;
int index;
scanf("%d",&N);
if(N>=1 && N <=50){
char temp[N];
double length,counter1;
double result[N];
for(i=0;i<N;i++){
fflush(stdin);
scanf("%s",&temp[i]);
}

for(i = 0;i<N;i++){
length=0;
counter1 = 0;
result[i] = 0;
length = strlen(temp[i]);
for(j = 0;j<length;j++){
if(temp[i][j] == '1'){
counter1++;
}
}
result[i] = counter1/length;

if(i==0){
tempVal = result;
index = 0;
}

if(tempVal > result[i]){
tempVal = result[i];
index = i;
}
}
printf("%d\n",index);
}
return 0;
}
```
Programming Problem Set #Elections

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