POJ-2042(最多四个数的平方和多少种方法)(Lagrange's Four-Square Theorem )
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AC的方法:
int d[400000];int ans[4000000];int main(){int i, j, k, l;for (i = 1; i <= 181; ++i) {d[i] = i * i;ans[d[i]]++;}for (i = 1; i <= 181; ++i) {for (j = i; j <= 181 && d[i] + d[j] <= 32768; ++j) {ans[d[i] + d[j]]++;}}for (i = 1; i <= 181; ++i) {for (j = i; j <= 181 && d[i] + d[j] <= 32768; ++j) {for (k = j; k <= 181 && d[i] + d[j] + d[k] <= 32768; ++k) {ans[d[i] + d[j] + d[k]]++;}}}for (i = 1; i <= 181; ++i) {for (j = i; j <= 181 && d[i] + d[j] <= 32768; ++j) {for (k = j; k <= 181 && d[i] + d[j] + d[k] <= 32768; ++k) {for (l = k; l <= 181 && d[i] + d[j] + d[k] + d[l] <= 32768; ++l) {ans[d[i] + d[j] + d[k] + d[l]]++;}}}}int n;while (cin>>n, n) {cout<<ans[n]<<endl;} return 0;}超时的方法:
int main(){int n;while (cin>>n, n) {int x = (int)(sqrt(n * 1.0) + 0.5);int i, ans = 0;int j, k, l, m;for (i = 1; i <= x; ++i) {if (i * i == n) ++ans;if (i * i > n) break;for (j = i; j <= x; ++j) {if (i * i + j * j == n) ++ans;if (i * i + j * j > n) break;for (k = j; k <= x; ++k) {if (i * i + j * j + k * k == n) ++ans;if (i * i + j * j + k * k > n) break;for (l = k; l <= x; ++l) {if (i * i + j * j + k * k + l * l == n) ++ans;if (i * i + j * j + k * k + l * l > n) break;}}}}printf("%d\n", ans);} return 0;}
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