1.
Sample Input
9
5 2 1 5 2 1 5 2 1
4
1 2 3 4
0
Sample Output
6
5
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[100];
bool vis[100];
int ans;
int f;
int n;
int sum;
bool cmp(int a, int b) {
return a > b;
}
void dfs(int x, int len, int g, int num) {
//当前下标 当前长度 目标长度 已匹配数量
//cout << "1x= " << x << " len = " << len << " g=" << g << " num=" << num << endl;
if(f) return;
if(num == n) {
f = 1;
return;
}
for(int i = x; i < n; ++i) {
if(!vis[i] && a[i] + len <= g) {
vis[i] = 1;
if(a[i] + len == g) {
//cout << "2x= " << x << " len = " << len << " g=" << g << " num=" << num << endl;
dfs(0, 0, g, num + 1);
}
else {
//cout << "3x= " << x << " len = " << len << " g=" << g << " num=" << num << endl;
dfs(i + 1, a[i] + len, g, num + 1);
}
//cout << 666 << endl;
vis[i] = 0; //去除标记
if(f) return;
if(len == 0) return; //剪枝
while(i < n && a[i+1] == a[i]) ++i; //剪枝
}
}
}
int main () {
while(cin >> n) {
//memset(vis, 0, sizeof vis);
if(!n) {
break;
}
sum = 0;
memset(a, 0, sizeof a);
for(int i = 0; i < n; ++i) {
cin >> a[i];
sum += a[i];
}
sort(a, a + n, cmp);
for(int i = a[0]; i <= sum; ++i) {
if(i == sum) {
cout << sum << endl;
}
else if(sum % i == 0) {//剪枝
f = 0;
dfs(0, 0, i, 0);
if(f == 1) {
cout << i << endl;
break;
}
}
}
}
}
2.
Sample Input
1
8
5
0
Sample Output
1
92
10
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int vis[3][50], P[15];//三个方向 ↖↑↗在此三个方向都不能有皇后
int n, sum;
void DFS(int row){
int i;
if (row == n + 1){//已经够n行了
sum++;
return;
}
for (i = 1; i <= n; i++){ // i表示第i列遍历 row表示第row行
// 关注对角线上数的行标和列标的特征
if (vis[0][row - i + n] == 0 && vis[1][i] == 0 && vis[2][row + i] == 0){//回溯
vis[0][row - i + n] = vis[1][i] = vis[2][row + i] = 1;//变值
DFS(row + 1);//深搜
vis[0][row - i + n] = vis[1][i] = vis[2][row + i] = 0;//回溯
}
}
}
int main()
{
for (n = 1; n <= 10; n++){//先打表不然会超时的
memset(vis, 0, sizeof(vis));
sum = 0;
DFS(1);
P[n] = sum;
}
while (cin >> n, n){
cout << P[n] << endl;
}
return 0;
}
3
Sample Input
7 1 2 ? 6 ? 3 5 8
? 6 5 2 ? 7 1 ? 4
? ? 8 5 1 3 6 7 2
9 2 4 ? 5 6 ? 3 7
5 ? 6 ? ? ? 2 4 1
1 ? 3 7 2 ? 9 ? 5
? ? 1 9 7 5 4 8 6
6 ? 7 8 3 ? 5 1 9
8 5 9 ? 4 ? ? 2 3
Sample Output
7 1 2 4 6 9 3 5 8
3 6 5 2 8 7 1 9 4
4 9 8 5 1 3 6 7 2
9 2 4 1 5 6 8 3 7
5 7 6 3 9 8 2 4 1
1 8 3 7 2 4 9 6 5
2 3 1 9 7 5 4 8 6
6 4 7 8 3 2 5 1 9
8 5 9 6 4 1 7 2 3
#include <bits/stdc++.h>
using namespace std;
char temp;
int ma[11][11] = {0};
int num = 0;
struct dian
{
int x, y;
}a[100];
bool isOK(int k, int step) {
int x, y;
for(int i = 0; i < 9; ++i) {
if(ma[a[step].x][i] == k || ma[i][a[step].y] == k){
return 0;
}
}
//剪枝
//判断这个数所处的那个小九宫格里面有没有重复的数
x = (a[step].x) / 3 * 3;
y = (a[step].y) / 3 * 3;
//这步十分重要
for(int i = x; i < x + 3; ++i) {
for(int j = y; j < y + 3; ++j) {
if(ma[i][j] == k) {
return 0;
}
}
}
return 1;
}
void Ptu(int m[][11]) {
for(int i = 0; i < 9; ++i) {
for(int j = 0; j < 9; ++j) {
cout << m[i][j];
if(j != 8) {
cout << " ";
}
}
cout << endl;
}
}
void DFS(int step) {
//cout << step << endl;
if(step == num) {
// for(int i=0; i < 9; i++){
// for(int j=0; j < 8; j++) {
// cout << ma[i][j] << " ";
// }
// cout << ma[i][8] << endl;//直接在这里输出结果，要不然会发生可怕的事~
// }
Ptu(ma);
return; //需要在这里输出结果，否则会被还原
}
for(int i = 1; i <= 9; ++i) {
if(isOK(i, step)) {
ma[a[step].x][a[step].y] = i;
DFS(step + 1);
ma[a[step].x][a[step].y] = 0;
}
}
return;
}
int main () {
int t = 0;
while(cin >> temp){
num = 0;
if(temp == '?') {
a[num].x = 0;
a[num].y = 0;
num++;
ma[0][0] = 0;
}
else {
ma[0][0] = temp - '0';
}
//memset(a, 0, sizeof a);
for(int i = 0; i < 9; ++i) {
for(int j = 0; j < 9; ++j) {
if(i == 0 && j == 0) {
continue;
}
cin >> temp;
if(temp == '?') {
//ma[i][j] == 0;
a[num].x = i;
a[num].y = j;
num++;
ma[i][j] = 0;
}
else {
ma[i][j] = temp - '0';
}
}
}
if(t++) {
putchar(10);
}
DFS(0);
}
return 0;
}
4.
#include <bits/stdc++.h>
using namespace std;
int i = 0;
string sh;
int f() {
int mx = 0;
int temp = 0;
int len = sh.length();
while(i < len) {
char t = sh[i++];
if(t == '(') {
temp += f();
}
else if (t == ')') {
break;
}
else if(t == '|') {
mx = max(temp, mx);
temp = 0;
}
else {
temp++;
}
}
return max(mx, temp);
}
int main () {
cin >> sh;
cout << f() << endl;
}
5.
//这道题最后一组样例被T了 只得了75分
//拿到满分需要用树的直径优化，改日再写
//注意要标记初始点
#include <bits/stdc++.h> using namespace std; int lu[10000][10000]; bool vis[100000]; int sum = 0; int temp = 0; int n; void DFS(int x) { //cout << "dfs"<<x<<endl; //cout << "x:" << x << endl; // if(x == n + 1) { // //cout << 66 << endl; // sum = max(temp, sum); // temp = 0; // return; // } for(int i = 1; i <= n; ++i) { vis[x] = 1; int tt = x < i ? lu[x][i] : lu[i][x]; if(tt && !vis[i]){ vis[i] = 1; // cout << "temp:" << temp << endl; // cout << "tt:" << tt << endl; temp += tt; DFS(i); vis[i] = 0; sum = max(temp, sum); temp -= tt; } vis[x] = 0; } //cout << "fdfs"<<x<<endl; } int main () { cin >> n; for(int i = 0; i < n - 1; ++i) { int a,b,c; cin >> a >> b >> c; if(a > b) swap(a, b); lu[a][b] = max(lu[a][b], c); } for(int i = 1; i <= n; ++i) { //memset(vis, 0, sizeof vis); DFS(i); } int s = 0; for(int i = 1; i <= sum; i++) { s += (i + 10); } //cout << sum << endl; cout << s << endl; }