2021“MINIEYE杯”中国大学生算法设计超级联赛（4）1001. Calculus（思维）

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

签到题，这些级数都是不收敛的，因此必须每一项的系数都是0才可以。特别注意常数项的情况。至于代码可以先把每一项拆开判断。

#include <bits/stdc++.h>
using namespace std;
int main() {
	int t;
	cin >> t;
	while(t--) {
		string s;
		cin >> s;
		int len = s.size();
		int lst = 0;
		vector<string> v;
		for(int i = 0; i <= len; i++) {
			if(i == len || s[i] == '+') {
				v.push_back(s.substr(lst, i - lst));
				lst = i + 1;
			}
		}
		bool flag = 1;
		for(auto x : v) {

			int pos = x.find("/");
			if(pos != x.npos) {
				if(!(pos == 1 && x[0] == '0')) {
					flag = 0;
					break;
				}
				continue;
			}

			int pos1 = x.find("^");
			if(pos1 != x.npos) {
				if(!(pos1 == 1 && (x[0] == '0' || x[0] == '1'))) {
					// cout << "fuck";
					flag = 0;
					break;
				}
				continue;
			}
			int pos2 = x.find("sin");
			if(pos2 != x.npos) {
				if(!(pos2 == 1 && x[0] == '0')) {
					flag = 0;
					break;
				}
				continue;
			}
			int pos3 = x.find("cos");
			if(pos3 != x.npos) {
				if(!(pos3 == 1 && x[0] == '0')) {
					flag = 0;
					break;
				}
				continue;
			}
			int pos4 = x.find("x");
			if(pos4 != x.npos) {
				if(!(pos4 == 1 && x[0] == '0')) {
					flag = 0;
					break;
				}
				continue;
			}
			if(x.size() > 1 || x[0] != '0') {
				flag = 0;
				break;
			}
		}
		if(flag) puts("YES");
		else puts("NO");
	}
	return 0;
}