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#作业代码``` c++//3.3  path 路径#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 数字接力#include#include#includeusing namespace std;  char a[1005][32];  char p[1005][32];    int n = 0;  //自己写一个字符串比较函数  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");  }    //将有二个有序数列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;

}

```