对输入的一元多项式，进行同类项合并，并按指数降序排序，输出处理后的一元多项式。如输入： "7X^4-5X^6+3X^3",输出： "-5X^6+7X^4+3X^3".
#include <map>
#include <string>
#include <stdlib.h>
#include <iostream>
using namespace std;
void OrderPolynomial (char* InputString, char* OutputString)
{
map< int, int, greater<int> > m;
string input(InputString);
if(input[0] != '-') {
input = "+" + input + "+";
} else {
input = input + "+";
}
unsigned int len = input.length();
bool isCoeff = true;
bool isMinus = false;
int coefficient = 0, exponent = 0;
for(unsigned int i = 0; i < len; ++i) {
if(input[i] == '-') {
isCoeff = true;
isMinus = true;
}
if(input[i] == '+') {
isCoeff = true;
isMinus = false;
}
if(input[i] >= '0' && input[i] <= '9') {
if(isCoeff){
coefficient = coefficient * 10 + input[i] - '0';
} else {
exponent = exponent * 10 + input[i] - '0';
}
}
if(input[i] == 'X') {
isCoeff = false;
if(isMinus){
coefficient = coefficient * -1;
}
}
if( (input[i] == '+' || input[i] == '-') && i != 0 ) {
if(m.count(exponent) == 0){
m[exponent] = coefficient;
} else {
m[exponent] += coefficient;
}
coefficient = 0;
exponent = 0;
}
}
string output = "";
map< int, int, greater<int> >::iterator it;
for(it = m.begin(); it != m.end(); ++it) {
char cc[20], ee[20];
itoa(it->second, cc, 10);
itoa(it->first, ee, 10);
string c(cc);
string e(ee);
if( it->second > 0 && it == m.begin() ) {
output += (c + "X^" + e);
}
if( it->second > 0 && it != m.begin() ) {
output += ("+" + c + "X^" + e);
}
if(it->second < 0) {
output += (c + "X^" + e);
}
}
output.copy(OutputString, output.length(),0);
return;
}
给出一个二叉树的前序和中序，求其后序。
#include <iostream>
#include <string>
using namespace std;
typedef struct Node{
char c;
struct Node *left;
struct Node *right;
}Node, *PNode;
static string pre;
static string mid;
void creatTreeByPreMid(PNode &p, int i, int j, int len){
if (len <= 0) return;
p = new Node;
p->c = pre[i];
p->left = NULL;
p->right = NULL;
int m = mid.find_first_of(pre[i]);
creatTreeByPreMid(p->left, i + 1, j, m - j);
creatTreeByPreMid(p->right, i + 1 + (m - j), m + 1, len - 1 - (m - j));
}
void postTraverse(PNode p){
if (p) {
postTraverse(p->left);
postTraverse(p->right);
cout << p->c;
}
}
int main(){
while (cin >> pre >> mid) {
PNode root;
creatTreeByPreMid(root, 0, 0, pre.length());
postTraverse(root);
cout << endl;
}
return 0;
}
一排同学N个，出队K个后，其他同学身高满足T1 < T2 < … < Ti , Ti > Ti+1 > … > TK (1 <= i <= K)，求至少需要出队多少。输入首行表示同学个数N，第二行依次为每个同学的身高。
#include <iostream>
using namespace std;
int main(){
int n, sum = 0;
cin >> n;
int inc1[200], inc2[200], a[200];
for (int i = 1; i <= n; ++i)
cin >> a[i];
inc1[1] = 1;
// inc1[i] 从左到 a[i] 最长上升子序列
// 状态转移递推式： inc1[i] = max{ inc1[j] | (1 <= j < i) 且 a[i] > a[j] } + 1
for (int i = 2; i <= n; ++i) {
inc1[i] = 1;
for (int j = 1; j < i; ++j) {
if (a[i] > a[j] && inc1[j] + 1 > inc1[i])
inc1[i] = inc1[j] + 1;
}
}
inc2[n] = 1;
for (int i = n - 1; i >= 1; --i) {
inc2[i] = 1;
for (int j = n; j > i; --j) {
if (a[i] > a[j] && inc2[j] + 1 > inc2[i])
inc2[i] = inc2[j] + 1;
}
}
for (int i = 1; i <= n; ++i) {
if (inc1[i] + inc2[i] - 1 > sum)
sum = inc1[i] + inc2[i] - 1;
}
cout << n - sum;
return 0;
}
一个整数拆成2的幂的和，求可以拆分的种数，输入n，输出 f(n) % 1000000000。
#include <iostream>
using namespace std;
// 一个整数总可以拆分为 2 的幂的和, 给定一个整数，求有多少种拆分方法
int main() {
int *p = new int[1000001];
int n;
while (cin >> n) {
p[0] = p[1] = 1;
for (int i = 2; i <= n; ++i) {
if (i % 2)
p[i] = p[i - 1];
else
p[i] = (p[i - 1] + p[i / 2]) % 1000000000;
}
cout << p[n] << endl;
}
delete[] p;
return 0;
}
输入输出用字符串表示大数
#include <iostream>
#include <string>
using namespace std;
void str2num(string str, int num[]){
for (unsigned int i = 0; i < str.length(); ++i) {
num[str.length() - 1 - i] = str[i] - '0';
}
}
// 大数求和，数字以字符串形式输入不超过100位
int main(){
string a, b;
cin >> a >> b;
int na[101] = { 0 };
int nb[100] = { 0 };
str2num(a, na);
str2num(b, nb);
for (unsigned int i = 0; i < 100; ++i) {
na[i] += nb[i];
na[i + 1] += na[i] / 10;
na[i] %= 10;
}
int i = 100;
for (; i >= 0; --i) {
if (na[i] != 0) {
break;
}
}
for (; i >= 0; --i) {
cout << na[i];
}
return 0;
}
名字为小写字母，其漂亮度是其所有字母漂亮度的总和，每个字母的漂亮度范围在1到26之间，并且都不同，求最大漂亮度。
#include <iostream>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
using namespace std;
// 名字是小写字母，与漂亮度数字[1,26]里的值一一对应自己安排，求名字的最大漂亮度值。
int main() {
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
string str;
cin >> str;
map<char, int> m;
for (unsigned int i = 0; i < str.size(); ++i) {
if (m.count(str[i]) == 0) {
m[str[i]] = 1;
} else {
++m[str[i]];
}
}
map<char, int>::iterator it;
vector<int> ve;
for (it = m.begin(); it != m.end(); ++it) {
ve.push_back(it->second);
}
sort(ve.begin(), ve.end(), greater<int>());
int sum = 0;
for (unsigned int i = 0, j = 26; i < ve.size(); ++i,--j) {
sum += ve[i] * j;
}
cout << sum << endl;
}
return 0;
}
有n个人围成一圈，顺序排号。从第一个人开始报数（从1到3报数），凡报到3的人退出，问最后留下的那位是原来第几号。
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
while (cin >> n) {
vector<int> ve;
for (int i = 1; i <= n; ++i) {
ve.push_back(i);
}
int out = 0, count = 0, index = 0;
while (out < n - 1) {
if (ve[index] != 0) {
++count;
}
if (count == 3) {
ve[index] = 0;
count = 0;
++out;
}
++index;
if (index == n) {
index = 0;
}
}
for (unsigned int i = 0; i < ve.size(); ++i) {
if (ve[i] != 0) {
cout << ve[i] << endl;
break;
}
}
}
return 0;
}
原文地址：http://blog.csdn.net/thisinnocence/article/details/44729237
