参考代码：
#include
using namespace std;
int n;
int rot[15], ans[15] = {0}; //分别存储不同次幂对应的系数
void dfs(int now, int exp, int an){
if(now == n){
ans[exp] += an;
return;
}
dfs(now+1,exp+1, an);
dfs(now+1, exp, an*rot[now]);
}
int main(){
int root;
scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%d", &rot[i]);
rot[i] = -rot[i];
}
dfs(0,0,1); //后面相乘的参量初始化为1
for(int i = n-1; i >= 0; i--) printf("%d%s", ans[i], i == 0 ? "\n" : " ");
return 0;
} 
参考代码：
#include
using namespace std;
const int maxn = 10010;
const int INF = 0x3fffffff;
struct triple{
int pos1, pos2, pos3;
};
int seq1[maxn], seq2[maxn], seq3[maxn];
int dis(int p1, int p2, int p3){
int maxNum = max(max(seq1[p1], seq2[p2]), seq3[p3]);
int minNum = min(min(seq1[p1], seq2[p2]), seq3[p3]);
return (maxNum-minNum)*2;
}
int main(){
int n1, n2, n3;
scanf("%d %d %d", &n1, &n2, &n3);
for(int i = 0; i < n1; i++) scanf("%d", &seq1[i]);
for(int i = 0; i < n2; i++) scanf("%d", &seq2[i]);
for(int i = 0; i < n3; i++) scanf("%d", &seq3[i]);
sort(seq1, seq1+n1);
sort(seq2, seq2+n2);
sort(seq3, seq3+n3);
int p1 = 0, p2 = 0, p3 = 0;
int minD = INF;
triple ans;
while(1){
int maxNum = max(max(seq1[p1], seq2[p2]), seq3[p3]);
int minNum = min(min(seq1[p1], seq2[p2]), seq3[p3]);
if((maxNum-minNum)*2 <= minD){
minD = (maxNum-minNum)*2;
ans.pos1 = p1;
ans.pos2 = p2;
ans.pos3 = p3;
}
if(seq1[p1] == minNum){
p1++;
if(p1 == n1) break;
}
else if(seq2[p2] == minNum){
p2++;
if(p2 == n2) break;
}
else{
p3++;
if(p3 == n3) break;
}
}
//最后这部分调节很重要，即调节中间的数字使其最大但不影响总的最小距离
while(ans.pos1+1 < n1 && minD == dis(ans.pos1+1, ans.pos2, ans.pos3)) ans.pos1++;
while(ans.pos2+1 < n2 && minD == dis(ans.pos1, ans.pos2+1, ans.pos3)) ans.pos2++;
while(ans.pos3+1 < n3 && minD == dis(ans.pos1, ans.pos2, ans.pos3+1)) ans.pos3++;
printf("MinD(%d, %d, %d) = %d\n", seq1[ans.pos1], seq2[ans.pos2], seq3[ans.pos3], minD);
return 0;
} 
参考代码：
#include
using namespace std;
const int maxn = 10010;
struct student{
int id, score;
}stu[1010];
struct node{
int id;
int num = 0, totalS = 0;
}isRoot[maxn];
int n;
int father[maxn];
int exist[maxn] = {0};
vector ans;
void init(){
for(int i = 0; i < maxn; i++){
father[i] = i;
}
}
int findFather(int a){
int x = a;
while(a != father[a]){
a = father[a];
}
while(x != a){
int tmp = father[x];
father[x] = a;
x = tmp;
}
return a;
}
void Union(int a, int b){
int fa = findFather(a);
int fb = findFather(b);
if(fa < fb) father[fb] = fa;
else if(fb < fa) father[fa] = fb;
}
bool cmp(node a, node b){
if(a.totalS != b.totalS) return a.totalS > b.totalS;
else if(a.num != b.num) return a.num < b.num;
else return a.id < b.id;
}
int main(){
init();
int k, teammate, score;
scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%d %d", &stu[i].id, &k);
exist[stu[i].id] = 1;
for(int j = 0; j < k; j++){
scanf("%d", &teammate);
exist[teammate] = 1;
Union(stu[i].id, teammate);
}
scanf("%d", &stu[i].score);
}
for(int i = 0; i < n; i++){//只能统计分数信息
int f = findFather(stu[i].id);
isRoot[f].id = f;
isRoot[f].totalS += stu[i].score;
}
for(int i = 0; i < maxn; i++){//统计人数信息
if(exist[i]){
int f = findFather(i); //此处i就是学生id不要再调用.id了
isRoot[f].num++;
}
}
for(int i = 0; i < maxn; i++){
if(isRoot[i].num) ans.push_back(isRoot[i]);
}
sort(ans.begin(), ans.end(), cmp);
cout << ans.size() << endl;
for(int i = 0; i < ans.size(); i++) printf("%04d %d %d\n", ans[i].id, ans[i].num, ans[i].totalS);
return 0;
} 
参考代码：
#include
using namespace std;
const int maxn = 100010;
int n, money;
int item[maxn], coupon[maxn];
struct node{
int id1, id2; //分别代表商品编号和代金券编号
friend bool operator < (node a, node b){
return item[a.id1] - coupon[a.id2] > item[b.id1] - coupon[b.id2];
}
};
bool cmp(int a, int b){
return a > n;
}
int main(){
scanf("%d %d", &n, &money);
for(int i = 0; i < n; i++) scanf("%d", &item[i]);
for(int i = 0; i < n; i++) scanf("%d", &coupon[i]);
sort(item, item+n);
sort(coupon, coupon+n, cmp);
priority_queue q;
for(int i = 0; i < n; i++){
q.push(node{i, 0});
}
int cnt = 0;
while(!q.empty()){
node first = q.top();
q.pop();
if(money >= item[first.id1] - coupon[first.id2]){
money -= (item[first.id1]-coupon[first.id2]);
cnt++;
if(first.id2+1 < n) q.push(node{first.id1, first.id2+1}); //最后的id2用+1不要用++
}else break;
}
printf("%d %d\n", cnt, money);
return 0;
}