HDU-2054.A==B?(字符串简单处理)
这道题......被我各种姿势搞死的...
本题大意:给出两个数A和B,判断A和B是否相等,对应输出YES or NO。
本题思路:本题我有两种思路,第一种是直接去除前导零和后导零然后稍加处理比较字符串即可,第二种是找出每个字符串的 '.' 然后向两边搜索即可,下面给出第一种思路的代码,仅供参考,建议读者自行实现。
参考代码:
#include <cstdio>
#include <cstring>
using namespace std; const int maxn = 1e8;
int begina, beginb, enda, endb, point1, point2, len1, len2;
bool flag;
char a[maxn], b[maxn]; int main () {
while(~scanf("%s %s", a, b)) {
flag = true;
point1 = point2 = maxn;
len1 = strlen(a), len2 = strlen(b);
if(strchr(a, '.'))
point1 = strchr(a, '.') - a;
if(strchr(b, '.'))
point2 = strchr(b, '.') - b;
begina = -, beginb = -, enda = len1, endb = len2;
for(int i = ; i < len1; i ++)// find begina
if(a[i] == '' && i < point1) begina = i;
else break;
for(int i = ; i < len2; i ++)// find beginb
if(b[i] == '' && i < point2) beginb = i;
else break;
for(int i = len1 - ; i >= point1; i --)// find enda
if(a[i] == '') enda = i;
else if(a[i] == '.') enda = i;
else break;
for(int i = len2 -; i >= point2; i --)// find endb
if(b[i] == '') endb = i;
else if(b[i] == '.') endb = i;
else break;
int i = begina + ;
for(int j = beginb + ; i < enda; j ++)
if(a[i ++] != b[j]) {
printf("NO\n");
flag = false;
break;
}
if(flag) printf("YES\n");
}
return ;
}
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