題目鏈接:點擊這裏
題意:給出一個數每個數位數字的和,每個數位數字的平方和,求出符合條件的最小數字。這個數字的數位不能超過100位。
乍一看像構造,其實存在最有策略。可以用dp[i][j]表示數位和爲i,數位平方和爲j的最短數字長度,這樣就可以
#include <bits/stdc++.h>
using namespace std;
long long x, y;
int dp[905][8105];
vector <int> ans;
int main () {
memset (dp, -1, sizeof dp);
dp[0][0] = 0;
for (int i = 1; i <= 900; i++) {
for (int j = 1; j <= 8100; j++) {
for (int bit = 1; bit <= 9; bit++) if (bit <= i && bit*bit <= j) {
if (dp[i-bit][j-bit*bit] == -1) continue;
if (dp[i][j] == -1) {
dp[i][j] = dp[i-bit][j-bit*bit]+1;
}
else
dp[i][j] = min (dp[i-bit][j-bit*bit]+1, dp[i][j]);
}
}
}
while (cin >> x >> y) {
if (x > 900 || y-x > 72*100 || y > 81*100 || y < x || ((y-x)&1) || dp[x][y] == -1 || dp[x][y] > 100) {
printf ("No solution\n");
continue;
}
if (x == y) {
for (int i = 1; i <= x; i++) printf ("1");
printf ("\n");
continue;
}
ans.clear ();
while (x && y) {
for (int i = 1; i <= 9; i++) {
if (dp[x-i][y-i*i] == dp[x][y]-1) {
ans.push_back (i);
x -= i;
y -= i*i;
break;
}
}
}
int sz = ans.size ();
for (int i = 0; i < sz; i++) {
printf ("%d", ans[i]);
}
printf ("\n");
}
return 0;
}