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;
}