Gym 100801E Easy Arithmetic (思维题)
2016-08-08 14:05
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题目:传送门。(需要下载PDF)
题意:给定一个长度不超过1000的字符串表达式,向该表达式中加入'+'或'-',使得表达式的值最大,输出该表达式。
题解:比如300-456就改成300-4+56,遇到二位数以上的减数的情况就变成-首位+剩下的,这样会使得表达式值最大。
题意:给定一个长度不超过1000的字符串表达式,向该表达式中加入'+'或'-',使得表达式的值最大,输出该表达式。
题解:比如300-456就改成300-4+56,遇到二位数以上的减数的情况就变成-首位+剩下的,这样会使得表达式值最大。
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> using namespace std; string c; int main() { freopen("easy.in","r",stdin); freopen("easy.out","w",stdout); cin>>c; int len=c.length(); bool flag=0; for(int i=0;i<len;i++) { if(c[i]=='-'||c[i]=='+') flag=0; if(flag) putchar('+'); if(c[i]!='0') flag=0; if(c[i-1]=='-'&&i>=1) flag=1; putchar(c[i]); } puts(""); return 0; }
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