Wednesday 13 April 2016

Solution of URI 1160 :: Population Increase

   , ,  7 comments

Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.

/**Bismillahir Rahmanir Rahim.**/

#include <stdio.h>
int main()
{
    int n, c, m;
    int a, b;
    double ac, bc;
    scanf("%d", &n);
    for(m=1; m<=n; m++)
    {
        c = 0;
        scanf("%d %d %lf %lf", &a, &b, &ac, &bc);
        while(a <= b)
        {
            a *= (ac / 100.0) + 1.0;
            b *= (bc / 100.0) + 1.0;
            c++;
            if (c > 100)
            {
                printf("Mais de 1 seculo.\n");
                break;
            }
        }
        if (c <= 100)
            printf("%d anos.\n", c);
    }
    return 0;
}

7 comments:

  1. Unknown31 July 2016 at 20:00

    Hello, why the +1 in a *= (ac / 100.0) + 1.0;
    b *= (bc / 100.0) + 1.0;
    ?
    Is this the same as
    R += ((int)(( (0.01*G1)-(0.01*G2) )*PA)); ?

    ReplyDelete
  2. Anonymous27 September 2016 at 23:27

    bujte parlam na formula ta

    ReplyDelete
  3. Anonymous4 December 2017 at 21:24

    Can you tell me please what are the differences between your code and my code (given below) ?

    #include

    int main()
    {
    int t,pa,pb,i,count;
    double g1,g2;

    scanf("%d",&t);

    for(i=1;i<=t;i++)
    { count =0;
    scanf("%d %d %lf %lf",&pa,&pb,&g1,&g2);
    while(pa<=pb)
    {

    pa *= (g1/100.0)+1.0;
    pb *= (g2/100.0)+1.0;

    count++;
    if(count>100)
    {
    printf("Mais de 1 seculo.\n");
    break;
    }

    }

    if (count <= 100)
    printf("%d anos\n",count);
    }
    return 0;
    }

    ReplyDelete
  4. Unknown1 May 2018 at 11:26

    #include
    int main()
    {
    int pa,pb,count,i,n;
    double ga,gb;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
    scanf("%d %d %lf %lf",&pa,&pb,&ga,&gb);
    count=0;

    while(pa<=pb)
    {
    pa=(pa+pa*(ga/100.00));

    count++;
    pb=(pb+pb*(gb/100.00));

    if(pa>pb)

    {
    if(count>100)
    {
    printf("Mais de 1 seculo.\n");
    }
    else if(count<=100)
    {
    printf("%d anos.\n",count);
    }

    break;
    }
    }
    }
    return 0;
    }
    What is the problem of my code?
    Oj shows that it is "exceeded Time limit".
    Plz help me..

    ReplyDelete
  5. Unknown3 July 2018 at 13:26

    ami ei problem ta python diye korechi....amar output thik ache...but uri online judge etate error dekhacche..amar problem ta te error ta khuje pacchi na....kindly apni jodi ektu dekhten khub valo hoto....please..!!

    T=int(input())
    if(1<=T<=3000):
    count=0
    PAL=list()
    PBL=list()
    G1L=list()
    G2L=list()
    while(1):
    PA,PB,G1,G2=input().split()
    PAL.append(PA)
    PBL.append(PB)
    G1L.append(G1)
    G2L.append(G2)
    count+=1
    if(count==T):
    break
    for i in range(0,T):
    pcount=0
    pa=int(PAL[i])
    pb=int(PBL[i])
    g1=float(G1L[i])
    g2=float(G2L[i])
    if((100<=pa<=1000000)and(100<=pb<=1000000)and(pag2)and(0.1<=g1<=10.0)and(0.0<=g2<=10.0)):
    while(1):
    pa=pa+int((g1*pa)/100.0)
    pb=pb+int((g2*pb)/100.0)
    pcount+=1
    if(pa>=pb):
    break
    if(pcount>100):
    print("Mais de 1 seculo .")
    else:
    print("%i anos."%pcount)

    ReplyDelete
  6. Unknown29 July 2019 at 21:21

    bro why i got TLE in this program??
    #include
    using namespace std;


    int main()
    {
    int a,b;
    int i=0,n=0,s;
    double ga,gb;
    scanf("%d",&n);
    while(n>0)
    {
    scanf("%d%d%lf%lf",&a,&b,&ga,&gb);
    while(b>=a)
    {
    a+=a*ga/100;
    b+=b*gb/100;
    i++;
    }
    if(i>100)
    printf("Mais de 1 seculo.\n");
    else
    printf("%d anos.\n",i);
    n--;
    i=0;
    }
    return 0;
    }

    ReplyDelete
  7. Unknown16 April 2020 at 19:39

    Bro do you build this website alone or someone else??

    ReplyDelete

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