本文主要是介绍7.29 第四场 Calculus，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

Calculus

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 398 Accepted Submission(s): 214

Problem Description

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.

f(x)=∑i=1nfi(x)

S(x)=∑j=1xf(j)

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

F={C,Cx,Csinx,Ccosx,Csinx,Ccosx,Cx,Cx}

C is a constant integer ranging from 0 to 109.

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

大概题意

判断所给函数是否收敛

思路

数学问题。。。直接判断字符串里有没有0即可

代码

#include<iostream>

using namespace std;
int t;
string s;

int main() {
    cin >> t;
    getchar();
    while (t--) {
        getline(cin, s);
        bool flag = false;
        for (int i = 0; i < s.length(); ++i) {
            if (s[i] > '0' && s[i] <= '9') {
                cout << "NO\n";
                flag = true;
                break;
            }
        }
        if (!flag)
            cout << "YES\n";
    }
    return 0;
}

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