T1:长度为n的集合中哪个子集和为c
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#include <iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string.h>
using namespace std;
int a[10000],ans[10000],sum[10000],n,c;
void print(int n)
{
for (int i=1;i<n;i++)
printf("%d ",ans[i]);
exit(0);
}
void dfs(int x,int tot,int k)
{
if (tot>c) return ;
if (tot==c) print(k);
if (tot+sum[x]<c) return ;
for (int i=x;i<=n;i++)
{
ans[k]=a[i];
dfs(i+1,tot+a[i],k+1);
}
}
int main(void)
{
scanf("%d%d",&n,&c);
for (int i=1;i<=n;i++)
scanf("%d",&a[i]);
for (int i=n;i>0;i--)
sum[i]=sum[i+1]+a[i];
if (sum[1]>=c) dfs(1,0,1);
else printf("No Solution!");
return 0;
}
T2:分配工作求min
跟昨天的题一毛一样 但考试一定要检查自己
交没交上去/是否保存!!!
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#include <bits/stdc++.h>
using namespace std;
int a[110][110],n;
int f[110], g[110], minn=1e6, tmp;
bool vis[110];
bool check() {
tmp = 0;
for (int i = 0; i < n; i++) tmp += a[i][f[i]];
return tmp < minn;
}
void load() {
minn = tmp;
for (int i = 0; i < n; i++) g[i] = f[i];
}
void dfs(int id) {
if (id == n) {
if (check())
load();
return;
}
for (int i = 0; i < n; i++) {
if (vis[i] == 0) {
f[id] = i;
vis[i] = 1;
dfs(id + 1);
vis[i] = 0;
}
}
}
void print() {
printf("%d",minn);
//for (int i = 0; i < n; i++) printf("%d ", g[i] + 1);
printf("\n");
}
int main() {
scanf("%d",&n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
}
}
dfs(0);
print();
return 0;
}
T3:n个数和尽量接近c
其实直接输出c是很实惠的
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T4:由ABC构成的长度为n的序列中无重复的相邻子序列
//能打表就打表,另外注意三个一组判重,而非两个一组
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char A='A',B='B',C='C';
char x[27][3]={{A,B,A},{B,C,A},{A,C,A},{B,A,A},{C,B,A},{C,A,A},{A,A,A},{B,B,A},{C,C,A},{A,B,B},{B,C,B},{A,C,B},{B,A,B},{C,B,B},{C,A,B},{A,A,B},{B,B,B},{C,C,B},{A,B,C},{B,C,C},{A,C,C},{B,A,C},{C,B,C},{C,A,C},{A,A,C},{B,B,C},{C,C,C}};
char a[15];
bool b[6];
long long int ans;
int n;
bool pd(int t,int i){
if(a[t-1]==a[t-3]&&'A'+i==a[t-2]) return 0;
return 1;
}
void search(int t){
for(int i=0;i<3;i++){
if(pd(t,i)){
a[t]='A'+i;
if(t==n) ans++;
else search(t+1);
}
}
}
3,9,27,72,198,540,1476,4032,11016,30096,82224,224640