#作業代碼``` c++//3.3? path 路徑#include#include#include#include#includeusing namespace std;struct node{int v, a, b, c[2000];}p[50][50];int mmin(int a,int b){if (a > b) return b;else return a;}int fac(int x){? return x*x;? }int main(){int T, o = 0;cin >> T;while (T--){int n, m;cin >> n >> m;memset(p, 0, sizeof(p));for (int i0 = 1;i0 <= n;i0++)for (int j = 1;j <= m;j++)cin >> p[i0][j].v;p[1][1].a = fac(p[1][1].v);p[1][1].b = p[1][1].v; p[1][1].c[p[1][1].v] = fac(p[1][1].v); for (int j = 2;j <= m;j++){p[1][j].a = p[1][j - 1].a + fac(p[1][j].v);p[1][j].b = p[1][j - 1].b + p[1][j].v;p[1][j].c[p[1][j].b] = p[1][j].a;}for (int i1 = 2;i1 <= n;i1++){p[i1][1].a = p[i1 - 1][1].a + fac(p[i1][1].v);p[i1][1].b = p[i1 - 1][1].b + p[i1][1].v;p[i1][1].c[p[i1][1].b] = p[i1][1].a;}for (int i = 2;i <= n;i++)for (int j = 2;j <= m;j++)for (int k = 0;k < 2000;k++) {if (p[i][j - 1].c[k]){int f = k + p[i][j].v;if (p[i][j].c[f])? ? ? ? ? ? ? ? ? ? ? p[i][j].c[f] = mmin(? p[i][j].c[f], p[i][j - 1].c[k] + fac(p[i][j].v)? );elsep[i][j].c[f] = p[i][j - 1].c[k] + fac(p[i][j].v);}if (p[i - 1][j].c[k]){int f = k + p[i][j].v;if (p[i][j].c[f])p[i][j].c[f] = mmin(p[i][j].c[f], p[i - 1][j].c[k] + fac(p[i][j].v));? ? ? ? ? ? ? ? ? ? ? ? elsep[i][j].c[f] = p[i - 1][j].c[k] + fac(p[i][j].v);}}int ans = 1000000000;for (int i2 = 0;i2 < 2000;i2++)if (p[n][m].c[i2])ans = mmin(ans, (n + m - 1)*p[n][m].c[i2] - fac(i2));printf("Case #%d: %d\n", ++o, ans);}return 0;}`````` c++//4.4 szjl 數字接力#include#include#includeusing namespace std;? char a[1005][32];? char p[1005][32];? ? int n = 0;? //自己寫一個字符串比較函數? int strcmp_vv(char s[], char t[])? {? ? ? char r[200];? ? ? char r1[200];? ? ? strcpy(r, s);? ? ? strcat(r, t);? ? ? strcpy(r1, t);? ? ? strcat(r1, s);? ? ? int i = strcmp(r, r1);? ? ? return i;? }? ? void outv()? {? ? ? ? for (int i = 0;i < n;i++)? ? ? ? ? ? printf("%s",a[i]);? ? ? ? ? ? ? ? printf("\n");? }? ? //將有二個有序數列a[first...mid]和a[mid...last]合并。? ? void mergearray(char a[][32], int first, int mid, int last,char temp[][32])? {? ? ? ? int i = first, j = mid + 1;? ? ? int m = mid, n = last;? ? ? int k = 0;? ? ? while (i <= m && j <= n)? ? ? {? ? ? ? if (strcmp_vv(a[i], a[j])==1)? ? ? ? ? ? ? ? strcpy(temp[k++], a[i++]);? ? ? ? ? ? else? ? ? ? ? ? ? ? strcpy(temp[k++], a[j++]);? ? ? }? ? ? while (i <= m)? ? ? ? ? strcpy(temp[k++] , a[i++]);? ? ? while (j <= n)? ? ? ? ? strcpy(temp[k++] , a[j++]);? ? ? for (i = 0; i < k; i++)? ? ? ? ? strcpy(a[first + i] , temp[i]); cout<<" first "<> n;
for (int i = 0;i < n;i++)? ? cin >> a[i];
MergeSort(a,n);
outv();
return 0;
}
```
