Problem E. Easy Arithmetic

最新推荐文章于 2023-10-09 02:02:27 发布

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最新推荐文章于 2023-10-09 02:02:27 发布

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本文链接：https://blog.youkuaiyun.com/weixin_44145887/article/details/97685202

本文介绍了一种算法，用于处理包含加减运算的算术表达式，通过在合适的位置添加加号，使最终表达式的计算结果达到最大值。算法首先读取原始表达式，然后检查每个减号后的数字，如果数字超过一位，则在减号和数字间插入加号，以此来增加表达式的总值。

Problem E. Easy Arithmetic
Input file: easy.in
Output file: easy.out
Time limit: 2 seconds
Memory limit: 256 megabytes
Eva is a third-grade elementary school student. She has just learned how to perform addition and
subtraction of arbitrary-precision integers. Her homework is to evaluate some expressions. It is boring,
so she decided to add a little trick to the homework. Eva wants to add some plus and minus signs to the
expression to make its value as large as possible.
Input
The single line of the input file contains the original arithmetic expression. It contains only digits, plus
(‘+’) and minus (‘-’) signs.
The original expression is correct, that is:
• numbers have no leading zeroes;
• there are no two consecutive signs;
• the last character of the expression is a digit.
The length of the original expression does not exceed 1000 characters.
Output
Output a single line — the original expression with some plus and minus signs added. Output expression
must satisfy the same correctness constraints as the original one. Its value must be as large as possible.
Examples
easy.in easy.out
10+20-30 10+20-3+0
-3-4-1 -3-4-1
+10 +10

题目给的已经是正确的表达式，现要求它的值最大，也就是在遇到负号时，如果负号后面的数字超过十位，就往里面添加加号。
-3010          -3+0+10
-30010        -3+0+0+10
-3330          -3+330
    /**/
    #include<stdio.h>
    #include<string.h>
    int main()
    {
        freopen("easy.in","r",stdin);
        freopen("easy.out","w",stdout);
        int i,j,len;
        int a[1010]={0};
        char s[1010];
        scanf("%s",s);
        len=strlen(s);
        for(i=0;i<len;i++)
        {
            if(s[i]=='-')
            {
                if(i+2<len&&s[i+2]!='+'&&s[i+2]!='-')
                {
                    for(j=i+2;j<len;j++)
                    {
                        if(s[j]=='0')
                            a[j]=1;
                        else if(s[j]>='1'&&s[j]<='9')
                        {
                            a[j]=1;
                            break;
                        }
                        else
                            break;
                    }
                }
            }
        }
        for(i=0;i<len;i++)
        {
            if(a[i])
                printf("+");
            printf("%c",s[i]);
        }
        printf("\n");
        return 0;
    }