zoukankan      html  css  js  c++  java
  • HDU-2054 A==B?

    #include<stdio.h>
    #include<string.h>
    char n[100000], m[100000];
    int main()
    {
    int i, j, len_n, len_m, flag, len_last, len_late,doc, q, p, count, start;
    char ch;
    while(1)
    {
    memset(n, 0, sizeof(n));
    memset(m, 0, sizeof(m));
    flag = i = j = doc = count = start = 0;
    while(1)
    {
    if(scanf("%c", &ch) == EOF) return 0;
    if(ch == ' ' && start == 1) break;
    else if(ch != ' ' && ch != ' ' && flag == 0)
    {
    n[i++] = ch;
    }
    else if(ch == ' ' || ch == ' ')
    {
    flag = 1;
    start = 1;
    }
    else m[j++] = ch;

    }
    len_n = strlen(n);
    len_m = strlen(m);
    if((n[0] == '-' && m[0] != '-') || (n[0] != '-' && m[0] == '-'))
    {
    printf("NO ");
    continue;
    }
    for(i = n[0] != '-'? 0:1; i<len_n && n[i] - '0'== 0; i++);
    for(j = m[0] != '-'? 0:1; j<len_m && m[j] - '0'== 0; j++);
    len_n -= i;
    len_m -= j;
    len_last = len_n < len_m? len_n:len_m;
    len_late = len_n + len_m - len_last;
    {
    for(; count < len_last; i++, j++, count++)
    if(n[i] - '0' != m[j] - '0') break;
    else if(n[i] == '.') doc = 1;
    if(count >= len_late)
    {
    printf("YES ");
    }
    else
    {
    if(len_n > len_m)
    {
    if(doc != 1 && n[i] != '.') printf("NO ");
    else
    {
    for(i = n[i] == '.'? i+1: i; i<strlen(n); i++)
    if(n[i] - '0' != 0 ) break;
    if(i>=len_n) printf("YES ");
    else printf("NO ");
    }

    }
    else if(len_n < len_m)
    {
    if(doc != 1 && m[j] != '.') printf("NO ");
    else
    {
    for(j = m[j] == '.'? j+1: j; j<strlen(m); j++)
    if(m[j] - '0' != 0 ) break;
    if(j>=len_m) printf("YES ");
    else printf("NO ");
    }
    }
    else if(len_n == len_m) printf("NO ");
    }
    }

    }
    return 0;
    }

    1.考虑数字前面的0

    2.考虑小数点后的0

    3.考虑大数字

  • 相关阅读:
    luoguP4389 付公主的背包 多项式exp
    bzoj3456 城市规划 多项式求In
    luoguP4491 [HAOI2018]染色 广义容斥原理 + FFT
    计蒜之道2019复赛题解
    Educational Codeforces Round 66 (Rated for Div. 2)
    [CodeChef-ANUDTQ] Dynamic Trees and Queries
    M-SOLUTIONS Programming Contest
    Codeforces Global Round 3
    PKUSC2019题解
    [LOJ#3120][Luogu5401][CTS2019]珍珠(容斥+生成函数)
  • 原文地址:https://www.cnblogs.com/52Cassie/p/4817496.html
Copyright © 2011-2022 走看看