1<=a<=9 0<=b,c<=9
我现在输入一个数n(n<=2000)
问 abc + acc = n 有多少种可能
输出要求：第一行一个数 x 代表可能的数量
接下来 x 行 ,每行输出 abc acc(如果有多组数组按照abc升序输出)
例子
输入
452
输出
1
241 211
#include <bits/stdc++.h>
#define maxn 1005
#define ll long long
using namespace std;
int a[maxn], b[maxn];
int main(){
int n, part1, part2;
while(scanf("%d",&n)!=EOF){
int ans = 0;
for(int i = 1 ; i <= 9; i++){
for(int j = 0; j <= 9; j++){
for(int k = 0; k <= 9; k++){
if(i == j || j == k || i == k) continue;
part1 = i * 100 + j * 10 + k;
part2 = i * 100 + k * 10 + k;
if(part1 + part2 == n){
a[ans] = part1;
b[ans++] = part2;
}
}
}
}
printf("%d\n",ans);
for(int i = 0; i< ans;i++){
printf("%d %d\n",a[i], b[i]);
}
}
return 0;
}
输出斐波那契数列输入一个 n 代表矩形 n x n
例子:
输入
3
输出
34 21 13
1 1 8
2 3 5
输入
4
输出
987 610 377 233
5 3 2 144
8 1 1 89
13 31 34 55
#include <bits/stdc++.h>
#define maxn 100
#define ll unsigned long long
using namespace std;
ll a[maxn];
int n;
void init(){
a[1] = 1, a[2] = 1;
for(int i = 3 ; i <= 81 ; i++){
a[i] = a[i - 1] + a[i - 2];
}
}
bool check(int x, int y){
if(x>=1 && x <= n && y >= 1 && y <= n)
return true;
return false;
}
ll ans[15][15];
bool mark[15][15];
int posi[][2] = {0,1,-1,0,0,-1,1,0};
int posiN[][2] = {0,-1,1,0,0,1,-1,0};
int main(){
init();
while(scanf("%d",&n)!=EOF){
memset(mark, false, sizeof(mark));
if(n % 2 == 0){
int beginX = n / 2 + 1;
int beginY = n / 2;
int pointer = 0;
int k = 1;
mark[beginX][beginY] = true;
ans[beginX][beginY] = 1;
while(beginX != 1 || beginY != 1){
beginX += posi[pointer][0];
beginY += posi[pointer][1];
if(mark[beginX][beginY] == false){
if(check(beginX, beginY)){
mark[beginX][beginY] = true;
ans[beginX][beginY] = a[++k];
pointer+=1;
pointer%=4;
}else{
pointer+=1;
pointer%=4;
}
}else{
beginX -= posi[pointer][0];
beginY -= posi[pointer][1];
pointer+=3;
pointer%=4;
}
}
}else{
int beginX = n / 2 + 1;
int beginY = n / 2 + 1;
int pointer = 0;
int k = 1;
mark[beginX][beginY] = true;
ans[beginX][beginY] = 1;
while(beginX != 1 || beginY != 1){
beginX += posiN[pointer][0];
beginY += posiN[pointer][1];
if(mark[beginX][beginY] == false){
if(check(beginX, beginY)){
mark[beginX][beginY] = true;
ans[beginX][beginY] = a[++k];
pointer+=1;
pointer%=4;
}else{
pointer+=1;
pointer%=4;
}
}else{
beginX -= posiN[pointer][0];
beginY -= posiN[pointer][1];
pointer+=3;
pointer%=4;
}
}
}
for(int i = 1 ; i <= n; i++){
for(int j = 1 ; j <= n; j++){
if(j == n){
cout<<ans[i][j]<<endl;
}else{
cout<<ans[i][j]<<" ";
}
}
}
}
return 0;
}
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