C 解法, 执行用时: 1ms, 内存消耗: 376KB, 提交时间: 2020-12-23

#include <stdio.h>
#include <string.h>
#define N 50
int main()
{
    char ch[N];
    scanf("%s",ch);
    int n,i,j,count=0;
    n=strlen(ch);
    int min=n;
    for(i=0;i<n;i++)
    {
        count=0;
        for(j=0;j<i;j++)
            if(ch[j]=='G')
                count++;
        for(j=i;j<n;j++)
            if(ch[j]=='R')
                count++;
        if(count<min)
            min=count;
    }
    printf("%d\n",min);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 348KB, 提交时间: 2020-10-29

#include <stdio.h>
#include <string.h>
#define N 50
int main()
{
    char ch[N];
    scanf("%s",ch);
    int n,i,j,count=0;
    n=strlen(ch);
    int min=n;
    for(i=0;i<n;i++)
    {
        count=0;
        for(j=0;j<i;j++)
            if(ch[j]=='G')
                count++;
        for(j=i;j<n;j++)
            if(ch[j]=='R')
                count++;
        if(count<min)
            min=count;
    }
    printf("%d\n",min);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2019-04-10

#include <stdio.h>
#include <string.h>
#define N 50
int main()
{
    char ch[N];
    scanf("%s",ch);
    int n,i,j,count=0;
    n=strlen(ch);
    int min=n;
    for(i=0;i<n;i++)
    {
        count=0;
        for(j=0;j<i;j++)
            if(ch[j]=='G')
                count++;
        for(j=i;j<n;j++)
            if(ch[j]=='R')
                count++;
        if(count<min)
            min=count;
    }
    printf("%d\n",min);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2019-08-31

#include <stdio.h>
#include <string.h>

/*--------------------------------------
We regard N(i) as the difference degree between 
the input string and string "RR...RGGG..G" which
contains i 'R's, We called this string as "i string".

Then we get N(i+1) = N(i) + d_i, d_i = 1 if i-th character of input
string is the same as the i-th character of "i string"
----------------------------------------*/

int Find_min(char* ipstr, int n)
{
    int cur = 0, cur_min;
    for (int i = 0; i < n; i++){
        if (*(ipstr+i) != 'G')
            cur++;
    }
    cur_min = cur;
    for (int i = 0; i < n; i++){
        if (*(ipstr+i) == 'G'){
            cur++;
        }
        else{
            cur--;
            if (cur < cur_min){
                cur_min = cur;
            }
        }
    }
    return cur_min;
}

int main(){
    char s[51];
    scanf("%s", s);
    printf("%d", Find_min(s, strlen(s)));
}

C 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2019-05-26

#include <stdio.h>
#include <string.h>
#define N 50
int main()
{
    char ch[N];
    scanf("%s",ch);
    int n,i,j,count=0;
    n=strlen(ch);
    int min=n;
    for(i=0;i<n;i++)
    {
        count=0;
        for(j=0;j<i;j++)
            if(ch[j]=='G')
                count++;
        for(j=i;j<n;j++)
            if(ch[j]=='R')
                count++;
        if(count<min)
            min=count;
    }
    printf("%d\n",min);
    return 0;
}