多校联赛四之多项式加法Calculus HDU - 6985（合并同类项）

This summer, ZXyang became so tired when doing the problems of Multi-University contests. So he decided to attend the Unified National Graduate Entrance Examination. This day, he sees a problem of series.

Let S(x) be a function with x as the independent variable. S(x) can be represented by the formula as follow.

fi(x) is a function with x as the independent variable. Furthermore. fi(x) belongs to the function set F.

C is a constant integer ranging from 0 to 10^9.

ZXyang wonders if S(x) is convergent. S(x) is convergent if and only if limx→∞S(x)=c, where c is a constant.

Input

The first line of input contains a single integer t (1≤t≤104) — the number of test cases.

The first and the only line of each test case contains a single string s (1≤|s|≤100), indicating the formula of f(x). Fraction is presented as a/b. Cx is presented as C^x. It’s guaranteed that the constant C won’t be left out when C=1. f(x) consists of functions from F connected with +.

Output

For each test case, print YES in one line if S(x) is a convergent sequence, or print NO in one line if not.

Sample Input

2
1sinx+0cosx+3x+6/sinx
0

Sample Output

NO
YES

题意

判断所给函数是否收敛

思路

题中所给的所有函数均为发散。所以只需要检查是否所有的构成函数的系数均为 0 即可

代码

#include<bits/stdc++.h>
using namespace std;
char a[1100],b[1100];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>a;
        int l=strlen(a);
        int c,f,k=0;
        int ans1=0,ans2=0,ans3=0,ans4=0,ans5=0,ans6=0,ans7=0,ans8=0;
        for(int i=0; i<l; i++)
        {
            if(a[i]!='+')
            {
                b[k++]=a[i];
                if(a[i+1]=='+'||a[i+1]=='\0')
                {
                    b[k]='\0';
                    c=0;
                    f=0;
                    for(int j=0; j<k; j++)
                    {
                        if(b[j]>='0'&&b[j]<='9')
                        {
                            c=c*10+b[j]-'0';
                        }
                        else
                        {
                            f=1;
                            if(b[j]=='/')
                            {
                                if(b[j+1]=='x')
                                    ans1+=c;
                                else if(b[j+1]=='s')
                                    ans2+=c;
                                else if(b[j+1]=='c')
                                    ans3+=c;
                            }
                            else if(b[j]=='x')
                                ans4+=c;
                            else if(b[j]=='s')
                                ans5+=c;
                            else if(b[j]=='c')
                                ans6+=c;
                            else if(b[j]=='^')
                                ans7+=c;
                            break;
                        }
                    }
                    if(f==0)
                        ans8+=c;
                    k=0;
                    c=0;
                }
            }
        }
        if(!ans1&&!ans2&&!ans3&&!ans4&&!ans5&&!ans6&&!ans7&&!ans8)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}