Codeforces Round #362 (Div. 2) B. Barnicle 科学记数法、表达式处理

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Codeforces Round #362 (Div. 2)

My Solution

处理科学计算法的表达式

找到 " . " 和 " e " 的位置， 然后读清题目的条件就好了

a and b contain no leading zeros and d contains no trailing zeros (but may be equal to 0). Also, b can not be non-zero if a is zero.

#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int maxn = 2*1e2 + 8;
char val[maxn], line[maxn] = {};
int main()
{
    #ifdef LOCAL
    freopen("a.txt", "r", stdin);
    //freopen("b.txt", "w", stdout);
    int T = 4;
    while(T--){
    #endif // LOCAL
    scanf("%s", val);
    int len = strlen(val), b;
    if(val[len-2] == 'e') b = val[len-1] - '0';
    else if(val[len-3] == 'e') b = 10*(val[len-2] - '0') + val[len-1] - '0';
    else b = 100;

    int cnti;
    for(int i = 1; i < len; i++){ if(b == 0) {cnti = 1; break;} if(val[i] != 'e'){ if(val[i+1] != 'e')val[i] = val[i+1]; else {val[i] = '.'; cnti = i; break;} b--; if(b == 0) {val[i+1] = '.'; cnti = i + 1; break;} } else break; } //printf("%sn %dn", val, cnti); bool _find = false, first = true; int ptr = 0; for(int i = len-1; i >= 0; i--){
        if(val[i] == 'e') _find = true;
        else if(_find){
            if(first && val[i] == '0') ;
            else{
                if(val[i] != '.')ptr = i;
                else ptr = i-1;
                break;
            }
        }
    }
    for(int i = 0; i <= ptr; i++)
        printf("%c", val[i]);
    /*
    bool first = true;
    int cnt0 = 0;
    for(int i = 0; i < len; i++){
        if(val[i] != 'e'){
            if(val[i] == '0' && first && i < cnti - 1) ;
            else if(val[i] == '0' && first && i == cnti - 1) {printf("0"); first = false;}
            else{
                if(first) first = false;
                if(!(val[i] == '.' && val[i+1] == 'e') &&  !(val[cnti] == '.' && val[cnti+1] == '0')){
                if(val[i] == '0'){
                    cnt0++;
                }
                if(val[i] != '0'){
                    while(cnt0--) printf("0");
                    printf("%c", val[i]);
                    cnt0 = 0;
                }

                }
                else if(val[cnti] == '.' && val[cnti+1] == '0'){
                    if(val[i] == '.') cnt0 = 1;
                    if(val[i] == '0'){
                        cnt0++;
                    }
                    if(val[i] != '0' && val[i] != '.'){
                        if(val[i-cnt0] == '.') {printf("."); cnt0--;}
                        while(cnt0--) printf("0");
                        printf("%c", val[i]);
                        val[]
                        cnt0 = 0;
                    }

                }
            }

        }
        else break;
    }
    */
    while(b) {printf("0"); b--;};

    #ifdef LOCAL
    printf("n");
    }
    #endif // LOCAL
    return 0;
}

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