CSP 202309-3 梯度求解

知识点1：stringsteam

stringsteam：支持对string 对象更灵活的处理;

作用:类型转换,词句转换

利用stringstream 实现句子与单个单词之间的转换；

1. 句子 转为单个的词

string a ="how old are you , dear ?";

stringstream ss;

ss << a;

string b;

while(ss >> b)

{

cout <<b<<endl;

}

字符串切割还可以使用如下方法：

getline(ss, 切割出来的子串, “间隔字符”);

while(getline(ss, tmp, ' ')){
        vec.push_back(tmp);
    }

2. 单个的词拼接为句子

stringstream ss;

ss << "how"<<" "<<"old"<<" "<<"are"<<" "<<"you"<<" "<<","<<"dear"<<" "<<"?";

string b;

b = ss.str();

cout<<b<<endl;

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
const int mod = 1e9 + 7;
using ll = long long;
ll coef[N];
int n, m, to_solve;
string s, tmp;
vector<string> vec;
void solve() {stack<map<ll, ll>> expr;for (auto& str: vec) {if (str.empty()) {continue;}if (str[0] == 'x') { // varll index = atoi(str.substr(1).c_str()); //取变量编号map<ll, ll> tp;if (index == to_solve) {	//如果是目标变量，标记 tp[1] = 1;	//使用0和1将目标变量和普通变量分隔开 } else {tp[0] = coef[index] % mod;	//不是，记录变量系数 }expr.push(tp);} else if (str.size() == 1 && !isdigit(str[0])) { //是操作符 auto pre = expr.top(); expr.pop();auto suf = expr.top(); expr.pop();map<ll, ll> res;if (str == "+") {for(auto& p: pre) {res[p.first] = (p.second % mod);	//提取变量系数 }for (auto& p: suf) {res[p.first] = (res[p.first] + p.second) % mod;}} else if (str == "-") {	//注意处理顺序 for(auto& p: suf) {res[p.first] = (p.second % mod);}for (auto& p: pre) {res[p.first] = (res[p.first] - p.second) % mod;}} else {for (auto& p: pre) {for (auto& q: suf) {ll zs = p.first + q.first;ll bs = (p.second * q.second) % mod;res[zs] = (res[zs] + bs) % mod;}}}expr.push(res);} else { // digitll digit = atoi(str.c_str());digit %= mod;map<ll, ll> tp;tp[0] = digit;expr.push(tp);}}assert(expr.size() == 1);	//如果expr大小为1，终止执行 ll res = 0;while (!expr.empty()) {auto final_expr = expr.top();expr.pop();for (auto& p: final_expr) {ll pw = 1;ll bs = (p.second * p.first) % mod;ll c = p.first - 1;while (c > 0) {c--;pw = (pw * coef[to_solve]) % mod;}pw = (pw * bs) % mod;res = (res + pw) % mod;}}res %= mod;res = (res + mod) % mod;cout << res << '\n';}
int main() {cin >> n >> m;getchar();	//回车 getline(cin, s);stringstream ss(s);while (getline(ss, tmp, ' ')) {vec.push_back(tmp);}for (int i = 1; i <= m; i++) {cin >> to_solve;for (int j = 1; j <= n; j++) {cin >> coef[j];}solve();}
}